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+\section{Secant of 30 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 30 \degrees = \sec \dfrac \pi 6 = \dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 30 \degrees
+      | r = \frac 1 {\cos 30 \degrees}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 3} 2}
+      | c = [[Cosine of 30 Degrees|Cosine of $30 \degrees$]]
+}}
+{{eqn | r = \frac {2 \sqrt 3} 3
+      | c = multiplying top and bottom by $2 \sqrt 3$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 45 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 45 \degrees = \sec \dfrac \pi 4 = \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 45 \degrees
+      | r = \frac 1 {\cos 45 \degrees}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 2} 2}
+      | c = [[Cosine of 45 Degrees|Cosine of $45 \degrees$]]
+}}
+{{eqn | r = \sqrt 2
+      | c = multiplying top and bottom by $2 \sqrt 2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 60 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 60 \degrees = \sec \dfrac \pi 3 = 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 60 \degrees
+      | r = \frac 1 {\cos 60 \degrees}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\frac 1 2}
+      | c = [[Cosine of 60 Degrees|Cosine of $60 \degrees$]]
+}}
+{{eqn | r = 2
+      | c = multiplying top and bottom by $2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 75 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 75 \degrees = \sec \dfrac {5 \pi} {12} = \sqrt 6 + \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 75 \degrees
+      | r = \frac 1 {\cos 75 \degrees}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 6 - \sqrt 2} 4}
+      | c = [[Cosine of 75 Degrees|Cosine of $75 \degrees$]]
+}}
+{{eqn | r = \frac 4 {\sqrt 6 - \sqrt 2}
+      | c = multiplying top and bottom by $4$
+}}
+{{eqn | r = \frac {4 \paren {\sqrt 6 + \sqrt 2} } {\paren {\sqrt 6 - \sqrt 2} \paren {\sqrt 6 + \sqrt 2} }
+      | c = multiplying top and bottom by $\sqrt 6 + \sqrt 2$
+}}
+{{eqn | r = \frac {4 \paren {\sqrt 6 + \sqrt 2} } {6 - 2}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \sqrt 6 + \sqrt 2
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Right Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 90 \degrees = \sec \dfrac \pi 2$ is undefined
+\end{theorem}
+
+\begin{proof}
+From [[Secant is Reciprocal of Cosine]]:
+:$\sec \theta = \dfrac 1 {\cos \theta}$
+From [[Cosine of Right Angle]]:
+:$\cos \dfrac \pi 2 = 0$
+Thus $\sec \theta$ is undefined at this value.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Angle plus Right Angle}
+Tags: Secant Function
+
+\begin{theorem}
+: $\sec \left({x + \dfrac \pi 2}\right) = -\csc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec \left({x + \frac \pi 2}\right)
+      | r = \frac 1 {\cos \left({x + \frac \pi 2}\right)} 
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {- \sin x}
+      | c = [[Cosine of Angle plus Right Angle]]
+}}
+{{eqn | r = -\csc x
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Angle plus Right Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+: $\csc \left({x + \dfrac \pi 2}\right) = \sec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc \left({x + \frac \pi 2}\right)
+      | r = \frac 1 {\sin \left({x + \frac \pi 2}\right)} 
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\cos x}
+      | c = [[Sine of Angle plus Right Angle]]
+}}
+{{eqn | r = \sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Complement equals Cosecant}
+Tags: Secant Function, Cosecant Function
+
+\begin{theorem}
+:$\map \sec {\dfrac \pi 2 - \theta} = \csc \theta$ for $\theta \ne n \pi$
+where $\sec$ and $\csc$ are [[Definition:Secant Function|secant]] and [[Definition:Cosecant|cosecant]] respectively.
+That is, the [[Definition:Cosecant|cosecant]] of an [[Definition:Angle|angle]] is the [[Definition:Secant Function|secant]] of its [[Definition:Complement of Angle|complement]].
+This relation is defined wherever $\sin \theta \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {\frac \pi 2 - \theta}
+      | r = \frac 1 {\map \cos {\frac \pi 2 - \theta} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\sin \theta}
+      | c = [[Cosine of Complement equals Sine]]
+}}
+{{eqn | r = \csc \theta
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+The above is valid only where $\sin \theta \ne 0$, as otherwise $\dfrac 1 {\sin \theta}$ is undefined.
+From [[Sine of Multiple of Pi]] it follows that this happens when $\theta \ne n \pi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Complement equals Secant}
+Tags: Secant Function, Cosecant Function
+
+\begin{theorem}
+:$\map \csc {\dfrac \pi 2 - \theta} = \sec \theta$ for $\theta \ne \paren {2 n + 1} \dfrac \pi 2$
+where $\csc$ and $\sec$ are [[Definition:Cosecant|cosecant]] and [[Definition:Secant Function|secant]] respectively.
+That is, the [[Definition:Secant Function|secant]] of an [[Definition:Angle|angle]] is the [[Definition:Cosecant|cosecant]] of its [[Definition:Complement of Angle|complement]].
+This relation is defined wherever $\cos \theta \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {\frac \pi 2 - \theta}
+      | r = \frac 1 {\map \sin {\frac \pi 2 - \theta} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\cos \theta}
+      | c = [[Sine of Complement equals Cosine]]
+}}
+{{eqn | r = \sec \theta
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+The above is valid only where $\cos \theta \ne 0$, as otherwise $\dfrac 1 {\cos \theta}$ is undefined.
+From [[Cosine of Half-Integer Multiple of Pi]] it follows that this happens when $\theta \ne \paren {2 n + 1} \dfrac \pi 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Supplementary Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\map \sec {\pi - \theta} = -\sec \theta$
+where $\sec$ denotes [[Definition:Secant Function|secant]].
+That is, the [[Definition:Secant Function|secant]] of an [[Definition:Angle|angle]] is the negative of its [[Definition:Supplement of Angle|supplement]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {\pi - \theta}
+      | r = \frac 1 {\map \cos {\pi - \theta} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {-\cos \theta} 
+      | c = [[Cosine of Supplementary Angle]]
+}}
+{{eqn | r = -\sec \theta
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Supplementary Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc \left({\pi - \theta}\right) = \csc \theta$
+where $\csc$ denotes [[Definition:Cosecant|cosecant]].
+That is, the [[Definition:Cosecant|cosecant]] of an [[Definition:Angle|angle]] equals its [[Definition:Supplement of Angle|supplement]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l=\csc \left({\pi - \theta}\right)
+      | r=\frac 1 {\sin \left({\pi - \theta}\right)}
+      | c=[[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r=\frac 1 {\sin \theta} 
+      | c=[[Sine of Supplementary Angle]]
+}}
+{{eqn | r=\csc \theta
+      | c=[[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Conjugate Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\map \sec {2 \pi - \theta} = \sec \theta$
+where $\sec$ denotes [[Definition:Secant Function|secant]].
+That is, the [[Definition:Secant Function|secant]] of an [[Definition:Angle|angle]] equals its [[Definition:Conjugate Angle|conjugate]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {2 \pi - \theta}
+      | r = \frac 1 {\map \cos {2 \pi - \theta} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\cos \theta}
+      | c = [[Cosine of Conjugate Angle]]
+}}
+{{eqn | r = \sec \theta
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Conjugate Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\map \csc {2 \pi - \theta} = -\csc \theta$
+where $\csc$ denotes [[Definition:Cosecant|cosecant]].
+That is, the [[Definition:Cosecant|cosecant]] of an [[Definition:Angle|angle]] is the negative of its [[Definition:Conjugate Angle|conjugate]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {2 \pi - \theta}
+      | r = \frac 1 {\map \sin {2 \pi - \theta} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {-\sin \theta}
+      | c = [[Cosine of Conjugate Angle]] and [[Sine of Conjugate Angle]]
+}}
+{{eqn | r = -\csc \theta
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 105 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 105 \degrees = \sec \dfrac {7 \pi} {12} = -\paren {\sqrt 6 + \sqrt 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 105 \degrees
+      | r = \map \sec {90 \degrees + 15 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 15 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -\paren {\sqrt 6 + \sqrt 2}
+      | c = [[Cosecant of 15 Degrees|Cosecant of $15 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 120 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 120 \degrees = \sec \dfrac {2 \pi} 3 = -2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 120 \degrees
+      | r = \map \sec {90 \degrees + 30 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 30 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -2
+      | c = [[Cosecant of 30 Degrees|Cosecant of $30 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 135 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 135 \degrees = \sec \dfrac {3 \pi} 4 = -\sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 135 \degrees
+      | r = \map \sec {90 \degrees + 45 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 45 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -\sqrt 2
+      | c = [[Cosecant of 45 Degrees|Cosecant of $45 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 150 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 150 \degrees = \sec \dfrac {5 \pi} 6 = -\dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 150 \degrees
+      | r = \map \sec {90 \degrees + 60 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 60 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -\dfrac {2 \sqrt 3} 3
+      | c = [[Cosecant of 60 Degrees|Cosecant of $60 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 165 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 165 \degrees = \sec \dfrac {11 \pi} {12} = -\paren {\sqrt 6 - \sqrt 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 165 \degrees
+      | r = \map \sec {90 \degrees + 75 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 75 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -\paren {\sqrt 6 - \sqrt 2}
+      | c = [[Cosecant of 75 Degrees|Cosecant of $75 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Straight Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 180 \degrees = \sec \pi = -1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 180 \degrees
+      | r = \map \sec {90 \degrees + 90 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 90 \degrees
+      | c = [[Secant of Angle plus Right Angle]]
+}}
+{{eqn | r = -1
+      | c = [[Cosecant of Right Angle]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 195 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 195 \degrees = \sec \dfrac {13 \pi} {12} = -\paren {\sqrt 6 - \sqrt 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 195 \degrees
+      | r = \map \sec {360^ \degrees - 165 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 165 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = -\paren {\sqrt 6 - \sqrt 2}
+      | c = [[Secant of 165 Degrees|Secant of $165 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 210 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 210 \degrees = \sec \dfrac {7 \pi} 6 = -2 \dfrac {\sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 210 \degrees
+      | r = \map \sec {360 \degrees - 150 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 150 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = -2 \frac {\sqrt 3} 3
+      | c = [[Secant of 150 Degrees|Secant of $150 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 225 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 225 \degrees = \sec \dfrac {5 \pi} 4 = -\sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 225 \degrees
+      | r = \map \sec {360 \degrees - 135 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 135 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = -\sqrt 2
+      | c = [[Secant of 135 Degrees|Secant of $135 degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 240 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 240 \degrees = \sec \dfrac {4 \pi} 3 = - 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 240 \degrees
+      | r = \map \sec {360 \degrees - 120 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 120 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = -2
+      | c = [[Secant of 120 Degrees|Secant of $120 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 255 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 255 \degrees = \sec \dfrac {17 \pi} {12} = -\paren {\sqrt 6 + \sqrt 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 255 \degrees
+      | r = \map \sec {360 \degrees - 105 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 105 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = -\paren {\sqrt 6 + \sqrt 2}
+      | c = [[Secant of 105 Degrees|Secant of $105 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Three Right Angles}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 270 \degrees = \sec \dfrac {3 \pi} 2$ is undefined
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 270 \degrees
+      | r = \map \sec {360 \degrees - 90 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 90 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{end-eqn}}
+But from [[Secant of Right Angle]], $\sec 90 \degrees$ is undefined.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 285 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 285 \degrees = \sec \dfrac {19 \pi} {12} = \sqrt 6 + \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 285 \degrees
+      | r = \map \sec {360 \degrees - 75 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 75 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = \sqrt 6 + \sqrt 2
+      | c = [[Secant of 75 Degrees|Secant of $75 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 300 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 300 \degrees = \sec \dfrac {5 \pi} 3 = 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 300 \degrees
+      | r = \map \sec {360 \degrees - 60 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 60 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = 2
+      | c = [[Secant of 60 Degrees|Secant of $60 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 315 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 315 \degrees = \sec \dfrac {7 \pi} 4 = \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 315 \degrees
+      | r = \map \sec {360 \degrees - 45 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 45 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = \sqrt 2
+      | c = [[Secant of 45 Degrees|Secant of $45 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 330 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 330 \degrees = \sec \dfrac {11 \pi} 6 = 2 \dfrac {\sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 330 \degrees
+      | r = \map \sec {360 \degrees - 30 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 30 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = 2 \frac {\sqrt 3} 3
+      | c = [[Secant of 30 Degrees|Secant of $30 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of 345 Degrees}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 345 \degrees = \sec \dfrac {23 \pi} {12} = \sqrt 6 - \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 345 \degrees
+      | r = \map \sec {360 \degrees - 15 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 15 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = \sqrt 6 - \sqrt 2
+      | c = [[Secant of 15 Degrees|Secant of $15 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Full Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\sec 360 \degrees = \sec 2 \pi = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec 360 \degrees
+      | r = \map \sec {360 \degrees - 0 \degrees}
+      | c = 
+}}
+{{eqn | r = \sec 0 \degrees
+      | c = [[Secant of Conjugate Angle]]
+}}
+{{eqn | r = 1
+      | c = [[Secant of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Zero}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 0$ is undefined
+\end{theorem}
+
+\begin{proof}
+From [[Cosecant is Reciprocal of Sine]]:
+: $\csc \theta = \dfrac 1 {\sin \theta}$
+From [[Sine of Zero is Zero]]:
+: $\sin 0 = 0$
+Thus $\csc \theta$ is undefined at this value.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 15 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 15^\circ = \csc \dfrac \pi {12} = \sqrt 6 + \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 15^\circ
+      | r = \frac 1 {\sin 15^\circ}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 6 - \sqrt 2} 4}
+      | c = [[Sine of 15 Degrees]]
+}}
+{{eqn | r = \frac 4 {\sqrt 6 - \sqrt 2}
+      | c = multiplying top and bottom by $4$
+}}
+{{eqn | r = \frac {4 \left({\sqrt 6 + \sqrt 2}\right)} {\left({\sqrt 6 - \sqrt 2}\right) \left({\sqrt 6 + \sqrt 2}\right)}
+      | c = multiplying top and bottom by $\sqrt 6 + \sqrt 2$
+}}
+{{eqn | r = \frac {4 \left({\sqrt 6 + \sqrt 2}\right)} {6 - 2}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \sqrt 6 + \sqrt 2
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 30 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 30^\circ = \csc \dfrac \pi 6 = 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 30^\circ
+      | r = \frac 1 {\sin 30^\circ}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\frac 1 2}
+      | c = [[Sine of 30 Degrees|Sine of $30^\circ$]]
+}}
+{{eqn | r = 2
+      | c = multiplying top and bottom by $2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 45 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 45^\circ = \csc \dfrac \pi 4 = \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 45^\circ
+      | r = \frac 1 {\sin 45^\circ}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 2} 2}
+      | c = [[Sine of 45 Degrees]]
+}}
+{{eqn | r = \sqrt 2
+      | c = multiplying top and bottom by $2 \sqrt 2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 60 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 60^\circ = \csc \dfrac \pi 3 = \dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 60^\circ
+      | r = \frac 1 {\sin 60^\circ}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 3} 2}
+      | c = [[Sine of 60 Degrees]]
+}}
+{{eqn | r = \frac {2 \sqrt 3} 3
+      | c = multiplying top and bottom by $2 \sqrt 3$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 75 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 75 \degrees = \csc \dfrac {5 \pi} {12} = \sqrt 6 - \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 75^\circ
+      | r = \frac 1 {\sin 75 \degrees}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\frac {\sqrt 6 + \sqrt 2} 4}
+      | c = [[Sine of 75 Degrees|Sine of $75 \degrees$]]
+}}
+{{eqn | r = \frac 4 {\sqrt 6 + \sqrt 2}
+      | c = multiplying top and bottom by $4$
+}}
+{{eqn | r = \frac {4 \paren {\sqrt 6 - \sqrt 2} } {\paren {\sqrt 6 + \sqrt 2} \paren {\sqrt 6 - \sqrt 2} }
+      | c = multiplying top and bottom by $\sqrt 6 - \sqrt 2$
+}}
+{{eqn | r = \frac {4 \paren {\sqrt 6 - \sqrt 2} } {6 - 2}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \sqrt 6 - \sqrt 2
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Right Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 90^\circ = \csc \dfrac \pi 2 = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 90^\circ
+      | r = \frac 1 {\sin 90^\circ}
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 1
+      | c = [[Sine of 90 Degrees]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 105 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 105 \degrees = \csc \dfrac {7 \pi} {12} = \sqrt 6 - \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 105 \degrees
+      | r = \map \csc {180 \degrees - 75 \degrees}
+      | c = 
+}}
+{{eqn | r = \csc 75 \degrees
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | r = \sqrt 6 - \sqrt 2
+      | c = [[Cosecant of 75 Degrees|Cosecant of $75 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 120 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 120 \degrees = \csc \dfrac {2 \pi} 3 = \dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 120 \degrees
+      | r = \map \csc {180 \degrees - 60 \degrees}
+      | c = 
+}}
+{{eqn | r = \csc 60 \degrees
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | r = \frac {2 \sqrt 3} 3
+      | c = [[Cosecant of 60 Degrees|Cosecant of $60 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 135 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 135 \degrees = \csc \dfrac {3 \pi} 4 = \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 135 \degrees
+      | r = \map \csc {180 \degrees - 45 \degrees}
+      | c = 
+}}
+{{eqn | r = \csc 45 \degrees
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | r = \sqrt 2
+      | c = [[Cosecant of 45 Degrees|Cosecant of $45 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 150 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 150 \degrees = \csc \dfrac {5 \pi} 6 = \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 150 \degrees
+      | r = \map \csc {180 \degrees - 30 \degrees}
+      | c = 
+}}
+{{eqn | r = \csc 30 \degrees
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | r = 2
+      | c = [[Cosecant of 30 Degrees|Cosecant of $30 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 165 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 165 \degrees = \csc \dfrac {11 \pi} {12} = \sqrt 6 + \sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 165 \degrees
+      | r = \map \csc {180 \degrees - 15 \degrees}
+      | c = 
+}}
+{{eqn | r = \csc 15 \degrees
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | r = \sqrt 6 + \sqrt 2
+      | c = [[Cosecant of 15 Degrees]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Straight Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 180^\circ = \csc \pi$ is undefined
+\end{theorem}
+
+\begin{proof}
+From [[Cosecant is Reciprocal of Sine]]:
+: $\csc \theta = \dfrac 1 {\sin \theta}$
+From [[Sine of Straight Angle]]:
+: $\sin \pi = 0$
+Thus $\csc \theta$ is undefined at this value.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 195 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 195^\circ = \csc \dfrac {13 \pi} {12} = - \left({\sqrt 6 + \sqrt 2}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 195^\circ
+      | r = \csc \left({360^\circ - 165^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 165^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = - \left({\sqrt 6 + \sqrt 2}\right)
+      | c = [[Cosecant of 165 Degrees]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 210 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 210^\circ = \csc \dfrac {7 \pi} 6 = -2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 210^\circ
+      | r = \csc \left({360^\circ - 150^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 150^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -2
+      | c = [[Cosecant of 150 Degrees|Cosecant of $150^\circ$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 225 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 225 \degrees = \csc \dfrac {5 \pi} 4 = -\sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 225 \degrees
+      | r = \map \csc {360 \degrees - 135 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 135 \degrees
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\sqrt 2
+      | c = [[Cosecant of 135 Degrees|Cosecant of $135 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 240 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 240 \degrees = \csc \dfrac {4 \pi} 3 = -\dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 240 \degrees
+      | r = \map \csc {360 \degrees - 120 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 120 \degrees
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\frac {2 \sqrt 3} 3
+      | c = [[Cosecant of 120 Degrees|Cosecant of $120 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 255 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 255^\circ = \csc \dfrac {17 \pi} {12} = -\left({\sqrt 6 - \sqrt 2}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 255^\circ
+      | r = \csc \left({360^\circ - 105^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 105^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\left({\sqrt 6 - \sqrt 2}\right)
+      | c = [[Cosecant of 105 Degrees|Cosecant of $105^\circ$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Three Right Angles}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 270^\circ = \csc \dfrac {3 \pi} 2 = -1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 270^\circ
+      | r = \csc \left({360^\circ - 90^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 90^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -1
+      | c = [[Cosecant of Right Angle]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 285 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 285^\circ = \csc \dfrac {19 \pi} {12} = -\left({\sqrt 6 - \sqrt 2}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 285^\circ
+      | r = \csc \left({360^\circ - 75^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 75^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\left({\sqrt 6 - \sqrt 2}\right)
+      | c = [[Cosecant of 75 Degrees]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 300 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 300^\circ = \csc \dfrac {5 \pi} 3 = -\dfrac {2 \sqrt 3} 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 300^\circ
+      | r = \csc \left({360^\circ - 60^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 60^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\frac {2 \sqrt 3} 3
+      | c = [[Cosecant of 60 Degrees]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 315 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 315^\circ = \csc \dfrac {7 \pi} 4 = -\sqrt 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 315^\circ
+      | r = \csc \left({360^\circ - 45^\circ}\right)
+      | c = 
+}}
+{{eqn | r = -\csc 45^\circ
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\sqrt 2
+      | c = [[Cosecant of 45 Degrees]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 330 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 330 \degrees = \csc \dfrac {11 \pi} 6 = -2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 330 \degrees
+      | r = \map \csc {360 \degrees - 30 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 30 \degrees
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -2
+      | c = [[Cosecant of 30 Degrees|Cosecant of $30 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of 345 Degrees}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 345 \degrees = \csc \dfrac {23 \pi} {12} = -\paren {\sqrt 6 + \sqrt 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc 345 \degrees
+      | r = \map \csc {360 \degrees - 15 \degrees}
+      | c = 
+}}
+{{eqn | r = -\csc 15 \degrees
+      | c = [[Cosecant of Conjugate Angle]]
+}}
+{{eqn | r = -\paren {\sqrt 6 + \sqrt 2}
+      | c = [[Cosecant of 15 Degrees|Cosecant of $15 \degrees$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Full Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\csc 360 \degrees = \csc 2 \pi$ is undefined
+\end{theorem}
+
+\begin{proof}
+From [[Cosecant is Reciprocal of Sine]]:
+:$\csc \theta = \dfrac 1 {\sin \theta}$
+From [[Sine of Full Angle]]:
+:$\sin 360 \degrees = 0$
+Thus $\csc \theta$ is undefined at this value.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Shape of Sine Function/Graph}
+Tags: Sine Function
+
+\begin{theorem}
+:[[File:Sine.png|800px]]
+\end{theorem}<|endoftext|>
+\section{Shape of Cosine Function/Graph}
+Tags: Cosine Function
+
+\begin{theorem}
+:[[File:Cosine.png|800px]]
+\end{theorem}<|endoftext|>
+\section{Shape of Tangent Function/Graph}
+Tags: Tangent Function
+
+\begin{theorem}
+:[[File:Tan.png|800px]]
+\end{theorem}<|endoftext|>
+\section{Shape of Cotangent Function/Graph}
+Tags: Cotangent Function
+
+\begin{theorem}
+:[[File:Cotangent.png|800px]]
+\end{theorem}<|endoftext|>
+\section{Shape of Secant Function/Graph}
+Tags: Secant Function
+
+\begin{theorem}
+:[[File:Secant.png|800px]]
+\end{theorem}<|endoftext|>
+\section{Cosecant Function is Odd}
+Tags: Cosecant Function, Examples of Odd Functions
+
+\begin{theorem}
+:$\map \csc {-x} = -\csc x$
+That is, the [[Definition:Cosecant|cosecant function]] is [[Definition:Odd Function|odd]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {-x}
+      | r = \frac 1 {\map \sin {-x} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {-\sin x}
+      | c = [[Sine Function is Odd]]
+}}
+{{eqn | r = -\csc x
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant Function is Even}
+Tags: Secant Function, Examples of Even Functions
+
+\begin{theorem}
+:$\map \sec {-x} = \sec x$
+That is, the [[Definition:Secant Function|secant function]] is [[Definition:Even Function|even]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {-x}
+      | r = \frac 1 {\map \cos {-x} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\cos x}
+      | c = [[Cosine Function is Even]]
+}}
+{{eqn | r = \sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \sech \paren {-x}
+	| r = \frac 1 {\cosh \paren {-x} }
+	| c = {{Defof|Hyperbolic Secant}}
+}}
+{{eqn	| r = \frac 1 {\cos \paren {-i x} }
+	| c = [[Hyperbolic Cosine in terms of Cosine]]
+}}
+{{eqn	| r = \frac 1 {\cos \paren {i x} }
+	| c = [[Cosine Function is Even]]
+}}
+{{eqn	| r = \frac 1 {\cosh x}
+	| c = [[Hyperbolic Cosine in terms of Cosine]]
+}}
+{{eqn	| r = \sech x
+	| c = {{Defof|Hyperbolic Secant}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sech {-x}
+      | r = \frac 1 {\map \cosh {-x} } 
+      | c = {{Defof|Hyperbolic Secant|index = 2}}
+}}
+{{eqn | r = \frac 1 {\cosh x} 
+      | c = [[Hyperbolic Cosine Function is Even]]
+}}
+{{eqn | r = \sech x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent Function is Odd}
+Tags: Cotangent Function, Examples of Odd Functions
+
+\begin{theorem}
+:$\map \cot {-x} = -\cot x$
+That is, the [[Definition:Cotangent|cotangent function]] is [[Definition:Odd Function|odd]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cot {-x}
+      | r = \frac {\map \cos {-x} } {\map \sin {-x} }
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{eqn | r = \frac {-\sin x} {\cos x}
+      | c = [[Cosine Function is Even]] and [[Sine Function is Odd]]
+}}
+{{eqn | r = -\cot x
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent of Sum}
+Tags: Cotangent Function
+
+\begin{theorem}
+:$\cot \left({a + b}\right) = \dfrac {\cot a \cot b - 1} {\cot b + \cot a}$
+where $\cot $ is [[Definition:Cotangent|cotangent]].
+=== [[Cotangent of Sum/Corollary|Corollary]] ===
+{{:Cotangent of Sum/Corollary}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cot \left({a + b}\right)
+      | r = \frac{\cos \left({a + b}\right)} {\sin \left({a + b}\right)} 
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{eqn | r = \frac {\cos a \cos b - \sin a \sin b} {\sin a \cos b + \cos a \sin b}
+      | c = [[Cosine of Sum]] and [[Sine of Sum]]
+}}
+{{eqn | r = \frac {\frac {\cos a \cos b} {\sin a \sin b} - 1} {\frac {\cos b} {\sin b} + \frac {\cos a} {\sin a} }
+      | c = dividing the [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] by $\sin a \sin b$
+}}
+{{eqn | r = \frac {\cot a \cot b - 1} {\cot b + \cot a}
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Half-Integer is Half Odd Integer}
+Tags: Integers, Odd Integers, Numbers
+
+\begin{theorem}
+Let $r$ be a [[Definition:Number|number]].
+Then $r$ is a [[Definition:Half-Integer|half-integer]] {{iff}} $r = \dfrac n 2$ where $n$ is an [[Definition:Odd Integer|odd integer]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $r$ be a [[Definition:Half-Integer|half-integer]].
+Then by definition $r = n + \dfrac 1 2$ for some $n \in \Z$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = 2 r
+      | r = 2 \paren {n + \dfrac 1 2}
+      | c = 
+}}
+{{eqn | r = 2 n + 2 \paren {\frac 1 2}
+      | c = 
+}}
+{{eqn | r = 2 n + 1
+      | c = 
+}}
+{{end-eqn}}
+thus showing that $r$ is half of $2 n + 1$ for some $n \in \Z$.
+By [[Odd Integer 2n + 1]] it follows that $r$ is half of an [[Definition:Odd Integer|odd integer]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $k$ be an [[Definition:Odd Integer|odd integer]].
+Then by [[Odd Integer 2n + 1]]:
+:$k = 2 n + 1$
+where $n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac k 2
+      | r = \frac {2 n + 1} 2
+      | c = 
+}}
+{{eqn | r = \frac {2 n} 2 + \frac 1 2
+      | c = 
+}}
+{{eqn | r = n + \frac 1 2
+      | c = 
+}}
+{{end-eqn}}
+thus showing that $\dfrac k 2$ is a [[Definition:Half-Integer|half-integer]].
+{{qed}}
+[[Category:Integers]]
+[[Category:Odd Integers]]
+[[Category:Numbers]]
+5mfab4beguz5qevypjz3d2aboc2jh8x
+\end{proof}<|endoftext|>
+\section{Sine of Angle plus Straight Angle}
+Tags: Sine Function
+
+\begin{theorem}
+:$\map \sin {x + \pi} = -\sin x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {x + \pi}
+      | r = \sin x \cos \pi + \cos x \sin \pi
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \sin x \cdot \paren {-1} + \cos x \cdot 0
+      | c = [[Cosine of Straight Angle]] and [[Sine of Straight Angle]]
+}}
+{{eqn | r = -\sin x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Angle plus Straight Angle}
+Tags: Cosine Function, Cosine of Angle plus Straight Angle
+
+\begin{theorem}
+:$\map \cos {x + \pi} = -\cos x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos \paren {x + \pi}
+      | r = \cos x \cos \pi - \sin x \sin \pi
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \cos x \cdot \paren {-1} - \sin x \cdot 0
+      | c = [[Cosine of Straight Angle]] and [[Sine of Straight Angle]]
+}}
+{{eqn | r = -\cos x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {x + \pi}
+      | r = \map \Re {\map \cos {x + \pi} + i \, \map \sin {x + \pi} }
+      | c = 
+}}
+{{eqn | r = \map \Re {e^{i \paren {x + \pi} } }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \map \Re {e^{i x + i \pi} }
+      | c = 
+}}
+{{eqn | r = \map \Re {e^{i x} e^{i \pi} }
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
+}}
+{{eqn | r = \map \Re {-e^{i x} }
+      | c = [[Euler's Identity]]
+}}
+{{eqn | r = -\map \Re {e^{i x} }
+      | c = 
+}}
+{{eqn | r = -\map \Re {\cos x + i \cos x}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = -\cos x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {x + \pi}
+      | r = \frac 1 2 \paren {e^{i \paren {x + \pi} } + e^{-i \paren {x + \pi} } }
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 2 \paren {e^{i x} e^{i \pi} + e^{-i x} e^{-i \pi} }
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
+}}
+{{eqn | r = \frac 1 2 \paren {-e^{i x} - e^{-i x} }
+      | c = [[Euler's Identity]]
+}}
+{{eqn | r = -\frac 1 2 \paren {e^{i x} + e^{-i x} }
+      | c = 
+}}
+{{eqn | r = -\cos x
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Angle plus Full Angle}
+Tags: Sine Function
+
+\begin{theorem}
+:$\map \sin {x + 2 \pi} = \sin x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {x + 2 \pi}
+      | r = \sin x \cos 2 \pi + \cos x \sin 2 \pi
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \sin x \cdot 1 + \cos x \cdot 0
+      | c = [[Cosine of Full Angle]] and [[Sine of Full Angle]]
+}}
+{{eqn | r = \sin x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Angle plus Full Angle}
+Tags: Cosine Function
+
+\begin{theorem}
+: $\cos \paren {x + 2 \pi} = \cos x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos \paren {x + 2 \pi}
+      | r = \cos x \cos 2 \pi - \sin x \sin 2 \pi
+      | c = [[cosine of Sum]]
+}}
+{{eqn | r = \cos x \cdot 1 - \sin x \cdot 0
+      | c = [[Cosine of Full Angle]] and [[Sine of Full Angle]]
+}}
+{{eqn | r = \cos x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent of Angle plus Full Angle}
+Tags: Tangent Function
+
+\begin{theorem}
+: $\tan \left({x + 2 \pi}\right) = \tan x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan \left({x + 2 \pi}\right)
+      | r = \frac {\sin \left({x + 2 \pi}\right)} {\cos \left({x + 2 \pi}\right)} 
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {\sin x} {\cos x}
+      | c = [[Sine of Angle plus Full Angle]] and [[Cosine of Angle plus Full Angle]]
+}}
+{{eqn | r = \tan x
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent of Angle plus Straight Angle}
+Tags: Tangent Function
+
+\begin{theorem}
+: $\tan \left({x + \pi}\right) = \tan x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan \left({x + \pi}\right)
+      | r = \frac {\sin \left({x + \pi}\right)} {\cos \left({x + \pi}\right)} 
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {-\sin x} {-\cos x}
+      | c = [[Sine of Angle plus Straight Angle]] and [[Cosine of Angle plus Straight Angle]]
+}}
+{{eqn | r = \tan x
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent of Angle plus Straight Angle}
+Tags: Cotangent Function
+
+\begin{theorem}
+:$\map \cot {x + \pi} = \cot x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cot {x + \pi}
+      | r = \frac {\map \cos {x + \pi} } {\map \sin {x + \pi} }
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{eqn | r = \frac {-\cos x} {-\sin x}
+      | c = [[Cosine of Angle plus Straight Angle]] and [[Sine of Angle plus Straight Angle]]
+}}
+{{eqn | r = \cot x
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Angle plus Straight Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\map \sec {x + \pi} = -\sec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {x + \pi}
+      | r = \frac 1 {\map \cos {x + \pi} } 
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {-\cos x}
+      | c = [[Cosine of Angle plus Straight Angle]]
+}}
+{{eqn | r = -\sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Angle plus Straight Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\map \csc {x + \pi} = -\csc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {x + \pi}
+      | r = \frac 1 {\map \sin {x + \pi} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {-\sin x}
+      | c = [[Sine of Angle plus Straight Angle]]
+}}
+{{eqn | r = -\csc x
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent of Angle plus Full Angle}
+Tags: Cotangent Function
+
+\begin{theorem}
+: $\cot \left({x + 2 \pi}\right) = \cot x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cot \left({x + 2 \pi}\right)
+      | r = \frac {\cos \left({x + 2 \pi}\right)} {\sin \left({x + 2 \pi}\right)}
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{eqn | r = \frac {\cos x} {\sin x}
+      | c =  [[Cosine of Angle plus Full Angle]] and [[Sine of Angle plus Full Angle]]
+}}
+{{eqn | r = \cot x
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Angle plus Full Angle}
+Tags: Secant Function
+
+\begin{theorem}
+: $\sec \left({x + 2 \pi}\right) = \sec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sec \left({x + 2 \pi}\right)
+      | r = \frac 1 {\cos \left({x + 2 \pi}\right)} 
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\cos x}
+      | c = [[Cosine of Angle plus Full Angle]]
+}}
+{{eqn | r = \sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Angle plus Full Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+: $\csc \left({x + 2 \pi}\right) = \csc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \csc \left({x + 2 \pi}\right)
+      | r = \frac 1 {\sin \left({x + 2 \pi}\right)} 
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {\sin x}
+      | c = [[Sine of Angle plus Straight Angle]]
+}}
+{{eqn | r = \csc x
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Angle plus Three Right Angles}
+Tags: Sine Function
+
+\begin{theorem}
+:$\map \sin {x + \dfrac {3 \pi} 2} = -\cos x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {x + \frac {3 \pi} 2}
+      | r = \sin x \cos \frac {3 \pi} 2 + \cos x \sin \frac {3 \pi} 2
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \sin x \cdot 0 + \cos x \cdot \paren {-1}
+      | c = [[Cosine of Three Right Angles]] and [[Sine of Three Right Angles]]
+}}
+{{eqn | r = -\cos x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Angle plus Three Right Angles}
+Tags: Cosine Function
+
+\begin{theorem}
+:$\map \cos {x + \dfrac {3 \pi} 2} = \sin x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {x + \frac {3 \pi} 2}
+      | r = \cos x \cos \frac {3 \pi} 2 - \sin x \sin \frac {3 \pi} 2
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \cos x \cdot 0 - \sin x \cdot \paren {-1}
+      | c = [[Cosine of Three Right Angles]] and [[Sine of Three Right Angles]]
+}}
+{{eqn | r = \sin x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent of Angle plus Three Right Angles}
+Tags: Tangent Function
+
+\begin{theorem}
+:$\map \tan {x + \dfrac {3 \pi} 2} = -\cot x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \tan {x + \frac {3 \pi} 2}
+      | r = \frac {\map \sin {x + \frac {3 \pi} 2} } {\map \cos {x + \frac {3 \pi} 2} } 
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {-\cos x} {\sin x}
+      | c = [[Sine of Angle plus Three Right Angles]] and [[Cosine of Angle plus Three Right Angles]]
+}}
+{{eqn | r = -\cot x
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent of Angle plus Three Right Angles}
+Tags: Cotangent Function
+
+\begin{theorem}
+:$\map \cot {x + \dfrac {3 \pi} 2} = -\tan x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cot {x + \frac {3 \pi} 2}
+      | r = \frac {\map \cos {x + \frac {3 \pi} 2} } {\map \sin {x + \frac {3 \pi} 2} }
+      | c = [[Cotangent is Cosine divided by Sine]]
+}}
+{{eqn | r = \frac {\sin x} {-\cos x}
+      | c = [[Cosine of Angle plus Right Angle]] and [[Sine of Angle plus Right Angle]]
+}}
+{{eqn | r = -\tan x
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Angle plus Three Right Angles}
+Tags: Secant Function
+
+\begin{theorem}
+:$\map \sec {x + \dfrac {3 \pi} 2} = \csc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {x + \frac {3 \pi} 2}
+      | r = \frac 1 {\map \cos {x + \frac {3 \pi} 2} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {\sin x}
+      | c = [[Cosine of Angle plus Three Right Angles]]
+}}
+{{eqn | r = \csc x
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Angle plus Three Right Angles}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\map \csc {x + \dfrac {3 \pi} 2} = -\sec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {x + \frac {3 \pi} 2}
+      | r = \frac 1 {\map \sin {x + \frac {3 \pi} 2} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {-\cos x}
+      | c = [[Sine of Angle plus Three Right Angles]]
+}}
+{{eqn | r = -\sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Three Right Angles less Angle}
+Tags: Sine Function
+
+\begin{theorem}
+:$\map \sin {\dfrac {3 \pi} 2 - \theta} = -\cos \theta$
+where $\sin$ and $\cos$ are [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {\frac {3 \pi} 2 - \theta}
+      | r = \sin \frac {3 \pi} 2 \cos \theta - \cos \frac {3 \pi} 2 \sin \theta
+      | c = [[Sine of Difference]]
+}}
+{{eqn | r = \paren {-1} \times \cos \theta - 0 \times \sin \theta
+      | c = [[Sine of Three Right Angles]] and [[Cosine of Three Right Angles]]
+}}
+{{eqn | r = -\cos \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Three Right Angles less Angle}
+Tags: Cosine Function
+
+\begin{theorem}
+:$\cos \left({\dfrac {3 \pi} 2 - \theta}\right) = - \sin \theta$
+where $\cos$ and $\sin$ are [[Definition:Cosine|cosine]] and [[Definition:Sine|sine]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l=\cos \left({\frac {3 \pi} 2 - \theta}\right)
+      | r=\cos \frac {3 \pi} 2 \cos \theta + \sin \frac {3 \pi} 2 \sin \theta
+      | c=[[Cosine of Difference]]
+}}
+{{eqn | r=0 \times \cos \theta + \left({-1}\right) \times \sin \theta
+      | c=[[Cosine of Three Right Angles]] and [[Sine of Three Right Angles]]
+}}
+{{eqn | r=- \sin \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Secant of Three Right Angles less Angle}
+Tags: Secant Function
+
+\begin{theorem}
+:$\map \sec {\dfrac {3 \pi} 2 - \theta} = -\csc \theta$
+where $\sec$ and $\csc$ are [[Definition:Secant Function|secant]] and [[Definition:Cosecant|cosecant]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sec {\frac {3 \pi} 2 - \theta}
+      | r = \frac 1 {\map \cos {\frac {3 \pi} 2 - \theta} }
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac 1 {-\sin \theta}
+      | c = [[Cosine of Three Right Angles less Angle]]
+}}
+{{eqn | r = -\csc \theta
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosecant of Three Right Angles less Angle}
+Tags: Cosecant Function
+
+\begin{theorem}
+:$\map \csc {\dfrac {3 \pi} 2 - \theta} = -\sec \theta$
+where $\csc$ and $\sec$ are [[Definition:Cosecant|cosecant]] and [[Definition:Secant Function|secant]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \csc {\frac {3 \pi} 2 - \theta}
+      | r = \frac 1 {\map \sin {\frac {3 \pi} 2 - \theta} }
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | r = \frac 1 {-\cos \theta}
+      | c = [[Sine of Three Right Angles less Angle]]
+}}
+{{eqn | r = -\sec \theta
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Angle Formulas/Sine}
+Tags: Sine Function, Double Angle Formula for Sine
+
+\begin{theorem}
+:$\sin 2 \theta = 2 \sin \theta \cos \theta$
+\end{theorem}<|endoftext|>
+\section{Double Angle Formulas/Cosine}
+Tags: Cosine Function, Double Angle Formula for Cosine
+
+\begin{theorem}
+:$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$
+\end{theorem}<|endoftext|>
+\section{Double Angle Formulas/Tangent}
+Tags: Tangent Function, Double Angle Formula for Tangent
+
+\begin{theorem}
+:$\map \tan {2 \theta} = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$
+\end{theorem}<|endoftext|>
+\section{Triple Angle Formulas}
+Tags: Trigonometric Identities, Hyperbolic Functions
+
+\begin{theorem}
+=== [[Triple Angle Formulas/Sine|Triple Angle Formula for Sine]] ===
+{{:Triple Angle Formulas/Sine}}
+=== [[Triple Angle Formulas/Cosine|Triple Angle Formula for Cosine]] ===
+{{:Triple Angle Formulas/Cosine}}
+=== [[Triple Angle Formulas/Tangent|Triple Angle Formula for Tangent]] ===
+{{:Triple Angle Formulas/Tangent}}
+\end{theorem}<|endoftext|>
+\section{Triple Angle Formulas/Sine}
+Tags: Sine Function, Triple Angle Formula for Sine
+
+\begin{theorem}
+:$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
+\end{theorem}<|endoftext|>
+\section{Triple Angle Formulas/Cosine}
+Tags: Cosine Function, Triple Angle Formula for Cosine
+
+\begin{theorem}
+:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
+\end{theorem}<|endoftext|>
+\section{Triple Angle Formulas/Tangent}
+Tags: Tangent Function, Triple Angle Formula for Tangent
+
+\begin{theorem}
+:$\map \tan {3 \theta} = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$
+\end{theorem}<|endoftext|>
+\section{Quadruple Angle Formulas}
+Tags: Trigonometric Identities, Hyperbolic Functions, Quadruple Angle Formulas
+
+\begin{theorem}
+=== [[Quadruple Angle Formulas/Sine|Quadruple Angle Formula for Sine]] ===
+{{:Quadruple Angle Formulas/Sine}}
+=== [[Quadruple Angle Formulas/Cosine|Quadruple Angle Formula for Cosine]] ===
+{{:Quadruple Angle Formulas/Cosine}}
+=== [[Quadruple Angle Formulas/Tangent|Quadruple Angle Formula for Tangent]] ===
+{{:Quadruple Angle Formulas/Tangent}}
+\end{theorem}<|endoftext|>
+\section{Quadruple Angle Formulas/Sine}
+Tags: Sine Function, Quadruple Angle Formula for Sine
+
+\begin{theorem}
+:$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$
+\end{theorem}<|endoftext|>
+\section{Quadruple Angle Formulas/Cosine}
+Tags: Cosine Function, Quadruple Angle Formulas, Quadruple Angle Formula for Cosine
+
+\begin{theorem}
+:$\cos 4 \theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$
+\end{theorem}<|endoftext|>
+\section{Quadruple Angle Formulas/Tangent}
+Tags: Tangent Function
+
+\begin{theorem}
+:$\tan 4 \theta = \dfrac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan 4 \theta
+      | r = \frac {\sin 4 \theta} {\cos 4 \theta}
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} {8 \cos^4 \theta - 8 \cos^2 \theta + 1}
+      | c = [[Quadruple Angle Formula for Sine]] and [[Quadruple Angle Formula for Cosine]]
+}}
+{{eqn | r = \frac {4 \frac {\tan \theta} {\cos^2 \theta} - 8 \tan^3 \theta} {8 - \frac 8 {\cos^2 \theta} + \frac 1 {\cos^4 \theta} }
+      | c = dividing top and bottom by $\cos^4 \theta$
+}}
+{{eqn | r =  \frac {4 \tan \theta \sec^2 \theta - 8 \tan^3 \theta} {8 - 8 \sec^2 \theta + \sec^4 \theta}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r =  \frac {4 \tan \theta \paren {1 + \tan^2 \theta} - 8 \tan^3 \theta} {8 - 8 \paren {1 + \tan^2 \theta} + \paren {1 + \tan^2 \theta}^2}
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r =  \frac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}
+      | c = multiplying out and gathering terms
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quintuple Angle Formulas/Sine}
+Tags: Sine Function, Quintuple Angle Formula for Sine
+
+\begin{theorem}
+:$\sin 5 \theta = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta$
+\end{theorem}<|endoftext|>
+\section{Quintuple Angle Formulas/Cosine}
+Tags: Cosine Function, Quintuple Angle Formula for Cosine
+
+\begin{theorem}
+:$\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$
+\end{theorem}<|endoftext|>
+\section{Quintuple Angle Formulas/Tangent}
+Tags: Tangent Function
+
+\begin{theorem}
+:$\tan 5 \theta = \dfrac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan 5 \theta
+      | r = \frac {\sin 5 \theta} {\cos 5 \theta}
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta} {16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta}
+      | c = [[Quintuple Angle Formulas]]
+}}
+{{eqn | r = \frac {5 \frac {\tan \theta} {\cos^4 \theta} - 20 \frac {\tan^3 \theta} {\cos^2 \theta} + 16 \tan^5 \theta} {16 \cos \theta - \frac {20} {\cos^2 \theta} + \frac 5 {\cos^4 \theta} }
+      | c = dividing top and bottom by $\cos^5 \theta$
+}}
+{{eqn | r = \frac {5 \tan \theta \sec^4 \theta - 20 \tan^3 \theta \sec^2 \theta + 16 \tan^5 \theta} {16 - 20 \sec^2 \theta + 5 \sec^4 \theta}
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac {5 \tan \theta \paren {1 + \tan^2 \theta}^2 - 20 \tan^3 \theta \paren {1 + \tan^2 \theta} + 16 \tan^5 \theta} {16 - 20 \paren {1 + \tan^2 \theta} + 5 \paren {1 + \tan^2 \theta}^2}
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r =  \frac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}
+      | c = multiplying out and gathering terms
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Sine Squared}
+Tags: Sine Function
+
+\begin{theorem}
+:$\sin^2 x = \dfrac {1 - \cos 2 x} 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1 - 2 \sin^2 x
+      | r = \cos 2 x
+      | c = [[Double Angle Formulas/Cosine/Corollary 2|Double Angle Formula for Cosine: Corollary 2]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sin^2 x
+      | r = \frac {\cos 2 x - 1} {-2}
+      | c = solving for $\sin^2x$
+}}
+{{eqn | r = \frac {1 - \cos 2 x} 2
+      | c = multiplying top and bottom by $-1$ and rearranging terms
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Cosine Squared}
+Tags: Cosine Function
+
+\begin{theorem}
+:$\cos^2 x = \dfrac {1 + \cos 2 x} 2$
+\end{theorem}<|endoftext|>
+\section{Power Reduction Formulas/Tangent Squared}
+Tags: Tangent Function
+
+\begin{theorem}
+:$\tan^2x = \dfrac {1 - \cos2x} {1 + \cos2x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan^2 x
+      | r = \frac {\sin^2 x} {\cos^2 x}
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {\frac {1 - \cos 2 x} 2} {\frac {\cos 2 x + 1} 2}
+      | c = [[Square of Sine]] and [[Square of Cosine]]
+}}
+{{eqn | r = \frac {1 - \cos 2 x} {1 + \cos 2 x}
+      | c = multiplying top and bottom by $2$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Tangent Function]]
+ct8rk66qffzsan69mylvbe9ebjskfb7
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Sine Cubed}
+Tags: Sine Function, Cube of Sine
+
+\begin{theorem}
+:$\sin^3 x = \dfrac {3 \sin x - \sin 3 x} 4$
+\end{theorem}<|endoftext|>
+\section{Power Reduction Formulas/Cosine Cubed}
+Tags: Cosine Function
+
+\begin{theorem}
+:$\cos^3 x = \dfrac {3 \cos x + \cos 3 x} 4$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos 3 x
+      | r = 4 \cos^3 x - 3 \cos x
+      | c = [[Triple Angle Formula for Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = 4 \cos^3 x
+      | r = 3 \cos x + \cos 3 x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \cos^3 x
+      | r = \dfrac {3 \cos x + \cos 3 x} 4
+      | c = dividing both sides by $4$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Sine to 4th}
+Tags: Sine Function
+
+\begin{theorem}
+:$\sin^4 x = \dfrac {3 - 4 \cos 2 x + \cos 4 x} 8$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin^4 x
+      | r = \paren {\sin^2 x}^2
+}}
+{{eqn | r = \paren {\frac {1 - \cos 2 x} 2}^2
+      | c = [[Square of Sine]]
+}}
+{{eqn | r = \frac {1 - 2 \cos 2 x + \cos^2 2 x} 4
+      | c = multiplying out
+}}
+{{eqn | r = \frac {1 - 2 \cos 2 x + \frac {1 + \cos 4 x} 2} 4
+      | c = [[Square of Cosine]]
+}}
+{{eqn | r = \frac {2 - 4 \cos 2 x + 1 + \cos 4 x} 8
+      | c = multiplying top and bottom by $2$
+}}
+{{eqn | r = \frac {3 - 4 \cos 2 x + \cos 4 x} 8
+      | c = rearrangement
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Cosine to 4th}
+Tags: Cosine Function, Power Reduction Formula for 4th Power of Cosine
+
+\begin{theorem}
+:$\cos^4 x = \dfrac {3 + 4 \cos 2 x + \cos 4 x} 8$
+\end{theorem}<|endoftext|>
+\section{Power Reduction Formulas/Sine to 5th}
+Tags: Sine Function
+
+\begin{theorem}
+:$\sin^5 x = \dfrac {10 \sin x - 5 \sin 3 x + \sin 5 x} {16}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin 5 x
+      | r = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x
+      | c = [[Quintuple Angle Formula for Sine]]
+}}
+{{eqn | ll= \leadsto
+      | l = 16 \sin^5 x
+      | r = \sin 5 x + 20 \sin^3 x - 5 \sin x
+      | c = rearranging
+}}
+{{eqn | r = \sin 5 x + 20 \paren {\frac {3 \sin x - \sin 3 x} 4} - 5 \sin x
+      | c = [[Power Reduction Formula for Cube of Sine]]
+}}
+{{eqn | r = \sin 5 x + 15 \sin x - 5 \sin 3 x - 5 \sin x
+      | c = multipying out
+}}
+{{eqn | r = 10 \sin x - 5 \sin 3 x + \sin 5 x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \sin^5 x
+      | r = \frac {10 \sin x - 5 \sin 3 x + \sin 5 x} {16}
+      | c = dividing both sides by 16
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas/Cosine to 5th}
+Tags: Cosine Function
+
+\begin{theorem}
+:$\cos^5 x = \dfrac {10 \cos x + 5 \cos 3 x + \cos 5 x} {16}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos 5 x
+      | r = 16 \cos^5 x - 20 \cos^3 x + 5 \cos x
+      | c = [[Quintuple Angle Formula for Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = 16 \cos^5 x
+      | r = \cos 5 x + 20 \cos^3 x - 5 \cos x
+      | c = rearranging
+}}
+{{eqn | r = \cos 5 x + 20 \paren {\frac {3 \cos x + \cos 3 x} 4} - 5 \cos x
+      | c = [[Power Reduction Formula for Cube of Sine]]
+}}
+{{eqn | r = \cos 5 x + 15 \cos x + 5 \cos 3 x - 5 \cos x
+      | c = multipying out
+}}
+{{eqn | r = 10 \cos x + 5 \cos 3 x + \cos 5 x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \cos^5 x
+      | r = \frac {10 \cos x + 5 \cos 3 x + \cos 5 x} {16}
+      | c = dividing both sides by 16
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Orthocomplement of Subset of Orthocomplement is Superset}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A, B \subseteq H$ be [[Definition:Subset|subsets]] of $H$ such that $B \subseteq A^\perp$, where $A^\perp$ is the [[Definition:Orthocomplement|orthocomplement]] of $A$.
+Then $A \subseteq B^\perp$.
+\end{theorem}
+
+\begin{proof}
+Let $B \subseteq A^\perp$.
+Then by [[Orthocomplement Reverses Subset]]:
+:$A^{\perp\perp} \subseteq B^\perp$
+By [[Double Orthocomplement is Closed Linear Span]] and the definition of [[Definition:Closed Linear Span|closed linear span]]:
+:$A \subseteq A^{\perp\perp}$
+Hence, by [[Subset Relation is Transitive]]:
+:$A \subseteq B^\perp$
+{{qed}}
+[[Category:Hilbert Spaces]]
+llqojzv5zg3rru2iy6w3z2fl6ud7645
+\end{proof}<|endoftext|>
+\section{Simpson's Formulas/Cosine by Cosine}
+Tags: Cosine Function, Simpson's Formulas
+
+\begin{theorem}
+:$\cos \alpha \cos \beta = \dfrac {\map \cos {\alpha - \beta} + \map \cos {\alpha + \beta} } 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \frac {\map \cos {\alpha - \beta} + \map \cos {\alpha + \beta} } 2
+}}
+{{eqn | r = \frac {\paren {\cos \alpha \cos \beta + \sin \alpha \sin \beta} + \paren {\cos \alpha \cos \beta - \sin \alpha \sin \beta} } 2
+      | c = [[Cosine of Difference]] and [[Cosine of Sum]]
+}}
+{{eqn | r = \frac {2 \cos \alpha \cos \beta} 2
+}}
+{{eqn | r = \cos \alpha \cos \beta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simpson's Formulas/Sine by Sine}
+Tags: Sine Function, Simpson's Formulas
+
+\begin{theorem}
+:$\sin \alpha \sin \beta = \dfrac {\map \cos {\alpha - \beta} - \map \cos {\alpha + \beta} } 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o =
+      | r = \frac {\map \cos {\alpha - \beta} - \map \cos {\alpha + \beta} } 2
+}}
+{{eqn | r = \frac {\paren {\cos \alpha \cos \beta + \sin \alpha \sin \beta} - \paren {\cos \alpha \cos \beta - \sin \alpha \sin \beta} } 2
+      | c = [[Cosine of Difference]] and [[Cosine of Sum]]
+}}
+{{eqn | r = \frac {2 \sin \alpha \sin \beta} 2
+}}
+{{eqn | r = \sin \alpha \sin \beta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simpson's Formulas/Cosine by Sine}
+Tags: Sine Function, Cosine Function, Simpson's Formulas, Simpson's Formula for Cosine by Sine
+
+\begin{theorem}
+:$\cos \alpha \sin \beta = \dfrac {\map \sin {\alpha + \beta} - \map \sin {\alpha - \beta} } 2$
+\end{theorem}<|endoftext|>
+\section{Simpson's Formulas/Sine by Cosine}
+Tags: Sine Function, Cosine Function, Simpson's Formulas, Simpson's Formulas: Sine by Cosine
+
+\begin{theorem}
+:$\sin \alpha \cos \beta = \dfrac {\map \sin {\alpha + \beta} + \map \sin {\alpha - \beta} } 2$
+\end{theorem}<|endoftext|>
+\section{Prosthaphaeresis Formulas/Sine plus Sine}
+Tags: Sine Function, Prosthaphaeresis, Compound Angle Formulas
+
+\begin{theorem}
+:$\sin \alpha + \sin \beta = 2 \map \sin {\dfrac {\alpha + \beta} 2} \map \cos {\dfrac {\alpha - \beta} 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 2 \map \sin {\dfrac {\alpha + \beta} 2} \map \cos {\dfrac {\alpha - \beta} 2}
+      | c = 
+}}
+{{eqn | r = 2 \frac {\map \sin {\dfrac {\alpha + \beta} 2 + \dfrac {\alpha - \beta} 2} + \map \sin {\dfrac {\alpha + \beta} 2 - \dfrac {\alpha - \beta} 2} } 2
+      | c = [[Simpson's Formula for Sine by Cosine]]
+}}
+{{eqn | r = \sin \frac {2 \alpha} 2 + \sin \frac {2 \beta} 2
+      | c = 
+}}
+{{eqn | r = \sin \alpha + \sin \beta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prosthaphaeresis Formulas/Sine minus Sine}
+Tags: Sine Function, Prosthaphaeresis, Compound Angle Formulas
+
+\begin{theorem}
+:$\sin \alpha - \sin \beta = 2 \map \cos {\dfrac {\alpha + \beta} 2} \map \sin {\dfrac {\alpha - \beta} 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 2 \map \cos {\frac {\alpha + \beta} 2} \map \sin {\frac {\alpha - \beta} 2}
+      | c = 
+}}
+{{eqn | r = 2 \frac {\map \sin {\dfrac {\alpha - \beta} 2 + \dfrac {\alpha + \beta} 2} + \map \sin {\dfrac {\alpha - \beta} 2 - \dfrac {\alpha + \beta} 2} } 2
+      | c = [[Simpson's Formula for Sine by Cosine]]
+}}
+{{eqn | r = \sin \frac {2 \alpha} 2 - \sin \frac {2 \beta} 2
+      | c = 
+}}
+{{eqn | r = \sin \alpha - \sin \beta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prosthaphaeresis Formulas/Cosine plus Cosine}
+Tags: Cosine Function, Prosthaphaeresis, Compound Angle Formulas
+
+\begin{theorem}
+:$\cos \alpha + \cos \beta = 2 \, \map \cos {\dfrac {\alpha + \beta} 2} \, \map \cos {\dfrac {\alpha - \beta} 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 2 \, \map \cos {\frac {\alpha + \beta} 2} \, \map \cos {\frac {\alpha - \beta} 2}
+      | c = 
+}}
+{{eqn | r = 2 \frac {\map \cos {\dfrac {\alpha + \beta} 2 - \dfrac {\alpha - \beta} 2} + \map \cos {\dfrac {\alpha + \beta} 2 + \dfrac {\alpha - \beta} 2} } 2
+      | c = [[Simpson's Formula for Cosine by Cosine]]
+}}
+{{eqn | r = \cos \frac {2 \beta} 2 + \cos \frac {2 \alpha} 2
+      | c = 
+}}
+{{eqn | r = \cos \alpha + \cos \beta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prosthaphaeresis Formulas/Cosine minus Cosine}
+Tags: Cosine Function, Prosthaphaeresis, Compound Angle Formulas
+
+\begin{theorem}
+:$\cos \alpha - \cos \beta = -2 \, \map \sin {\dfrac {\alpha + \beta} 2} \, \map \sin {\dfrac {\alpha - \beta} 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = -2 \, \map \sin {\frac {\alpha + \beta} 2} \, \map \sin {\frac {\alpha - \beta} 2}
+      | c = 
+}}
+{{eqn | r = -2 \frac {\map \cos {\dfrac {\alpha + \beta} 2 - \dfrac {\alpha - \beta} 2} - \map \cos {\dfrac {\alpha + \beta} 2 + \dfrac {\alpha - \beta} 2} } 2
+      | c = [[Simpson's Formula for Sine by Sine]]
+}}
+{{eqn | r = -\paren {\cos \frac {2 \beta} 2 - \cos \frac {2 \alpha} 2}
+      | c = 
+}}
+{{eqn | r = \cos \alpha - \cos \beta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Integer Multiple of Argument/Formulation 1}
+Tags: Sine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \sin n \theta
+      | r = \sin \theta \paren {\paren {2 \cos \theta}^{n - 1} - \dbinom {n - 2} 1 \paren {2 \cos \theta}^{n - 3} + \dbinom {n - 3} 2 \paren {2 \cos \theta}^{n - 5} - \cdots}
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sin \theta
+      | r = \sin \theta
+      | c = 
+}}
+{{eqn | r = \sin \theta \binom {1 - 1 } 0 \paren {2 \cos \theta}^{1 - \paren {0 + 1} }
+      | c = [[Zero Choose Zero]], {{Defof|Integer Power}}
+}}
+{{end-eqn}}
+So $\map P 1$ is seen to hold.
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sin 2 \theta
+      | r = 2 \sin \theta \cos \theta
+      | c = [[Double Angle Formulas/Sine]]
+}}
+{{eqn | r = \sin \theta \binom {2 - 1 } 0 \paren {2 \cos \theta}^{2 - \paren {0 + 1} }
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 2$ is also seen to hold.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \sin {n \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$
+from which we are to show:
+:$\displaystyle \map \sin {\paren {n + 1} \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+To proceed, we will require the following lemma:
+=== [[Sine of Integer Multiple of Argument/Formulation 1/Lemma|Lemma]] ===
+{{:Sine of Integer Multiple of Argument/Formulation 1/Lemma}}{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \map \sin {\paren {n + 1} \theta} 
+      | r = \map \sin {n \theta + \theta }
+      | c =
+}}
+{{eqn | r = \sin n \theta \cos \theta + \cos n \theta \sin \theta
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \sin n \theta \cos \theta + \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta} 
+      | c = [[Sine of Integer Multiple of Argument/Formulation 1/Lemma|Lemma above]]
+}}
+{{eqn | r = \sin n \theta \paren {2 \cos \theta } - \map \sin {\paren {n - 1 } \theta} 
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } \paren {2 \cos \theta } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n - 1 - \paren {2 k + 1} } }
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 2} } k \paren {2 \cos \theta}^{n - \paren {2 k + 2} } }
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k - 1 + 2} } {k - 1 } \paren {2 \cos \theta}^{n - \paren {2 \paren {k - 1 } + 2} } } }
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - \paren {k + 1} } k + \binom {n - \paren {k + 1} } {k - 1 } } \paren {2 \cos \theta}^{n - \paren {2 k} } } }
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }
+      | c = adding and subtracting $1$
+}}
+{{end-eqn}}
+The result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Integer Multiple of Argument}
+Tags: Cosine Function, Cosine of Integer Multiple of Argument
+
+\begin{theorem}
+For $n \in \Z_{>0}$:
+\end{theorem}<|endoftext|>
+\section{Cosine to Power of Odd Integer}
+Tags: Cosine Function, Cosine to Power of Odd Integer
+
+\begin{theorem}
+:$\displaystyle \cos^{2n+1} \theta = \frac 1 {2^{2n}} \sum_{k \mathop = 0}^n \binom {2n+1} k \cos \left({2n - 2k + 1}\right) \theta$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^{2 n + 1} \theta
+      | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^{2 n + 1}
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \paren {e^{i \theta} + e^{-i \theta} }^{2 n + 1}
+      | c = [[Exponent Combination Laws/Power of Product|Exponent Combination Laws: Power of Product]]
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \sum^{2 n + 1}_{k \mathop = 0} \binom {2 n + 1} k e^{k i \theta} e^{-\paren {2 n - k + 1} i \theta}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \sum^{2 n + 1}_{k \mathop = 0} \binom {2 n + 1} k e^{-\paren {2 n - 2 k + 1} i \theta}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \paren {\sum^n_{k \mathop = 0} \binom {2 n + 1} k e^{-\paren {2 n - 2 k + 1} i \theta} + \sum^{2 n + 1}_{k \mathop = n + 1} \binom {2 n + 1} k e^{-\paren {2 n - 2 k + 1} i \theta} }
+      | c = Partitioning the sum
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \paren {\sum^n_{k \mathop = 0} \binom {2 n + 1} k e^{-\paren {2 n - 2 k + 1} i \theta} + \sum^n_{k \mathop = 0} \binom {2 n + 1} {2 n + 1 - k} e^{\paren {2 \paren {2 n - k + 1} - 2 n - 1} i \theta} }
+      | c = $k \mapsto 2 n + 1 - k$
+}}
+{{eqn | r = \frac 1 {2^{2 n + 1} } \paren {\sum^n_{k \mathop = 0} \binom {2 n + 1} k e^{-\paren {2 n - 2 k + 1} i \theta} + \sum^n_{k \mathop = 0} \binom {2 n + 1} k e^{\paren {2 n - 2 k + 1} i \theta} }
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \sum^n_{k \mathop = 0} \binom {2 n + 1} k \, \map \cos {2 n - 2 k + 1} \theta
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^n \theta
+      | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^n
+      | c = [[De Moivre's Theorem]]
+}}
+{{eqn | r = \frac {\paren {e^{i \theta} + e^{-i \theta} }^n} {2^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - k} i \theta} e^{-k i \theta}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - 2 k} i \theta}
+      | c = 
+}}
+{{end-eqn}}
+Matching up terms from the beginning of this expansion with those from the end:
+{{begin-eqn}}
+{{eqn | l = 2^n \cos^n \theta
+      | r = e^{n i \theta} + \binom n 1 e^{\paren {n - 2} i \theta} + \binom n 2 e^{\paren {n - 4} i \theta} + \cdots
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \binom n {n - 2} e^{-\paren {n - 4} i \theta} + \binom n {n - 1} e^{-\paren {n - 2} i \theta} + e^{-n i \theta}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2 n} 2 \cos^n \theta
+      | r = \paren {\frac {e^{n i \theta} + e^{-n i \theta} } 2} + \binom n 1 \paren {\frac {e^{\paren {n - 2} i \theta} + e^{-\paren {n - 2} i \theta} } 2}
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \binom n 2 \paren {\frac {e^{\paren {n - 4} i \theta} + e^{-\paren {n - 4} i \theta} } 2} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$
+Now to determine $R_n$.
+The middle two terms of the sequence $0, 1, \ldots, n$ are $\dfrac {n - 1} 2$ and $\dfrac {n + 1} 2$.
+Thus, when $k = \dfrac {n - 1} 2$:
+{{begin-eqn}}
+{{eqn | l = n - 2 k
+      | r = n - 2 \frac {n - 1} 2
+      | c = 
+}}
+{{eqn | r = n - \frac {2 n - 2} 2
+      | c = 
+}}
+{{eqn | r = n - n + 1
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Similarly, when $k = \dfrac {n + 1} 2$:
+{{begin-eqn}}
+{{eqn | l = n + 2 k
+      | r = n - 2 \frac {n + 1} 2
+      | c = 
+}}
+{{eqn | r = n - \frac {2 n + 2} 2
+      | c = 
+}}
+{{eqn | r = n - n - 1
+      | c = 
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Binomial Coefficient|binomial coefficient]] in each case is the same, because:
+{{begin-eqn}}
+{{eqn | l = n - \frac {n - 1} 2
+      | r = \frac {2 n - n - 1} 2
+      | c = 
+}}
+{{eqn | r = \frac {n + 1} 2
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \binom n {\paren {n - 1} / 2}
+      | r = \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!}
+      | c = 
+}}
+{{eqn | r = \binom n {\paren {n + 1} / 2}
+      | c = 
+}}
+{{end-eqn}}
+Thus the two middle terms collapse to:
+{{begin-eqn}}
+{{eqn | l = R_n
+      | r = \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!} \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}
+      | c = 
+}}
+{{eqn | r = \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!} \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine to Power of Even Integer}
+Tags: Cosine Function, Cosine to Power of Even Integer
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \cos^{2 n} \theta
+      | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n - 1} } \paren {\cos 2 n \theta + \binom {2 n} 1 \map \cos {2 n - 2} \theta + \cdots + \binom {2 n} {n - 1} \cos 2 \theta}
+      | c = 
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \binom {2 n} k \map \cos {2 n - 2 k} \theta
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^n \theta
+      | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^n
+      | c = [[De Moivre's Theorem]]
+}}
+{{eqn | r = \frac {\paren {e^{i \theta} + e^{-i \theta} }^n} {2^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - k} i \theta} e^{-k i \theta}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - 2 k} i \theta}
+      | c = 
+}}
+{{end-eqn}}
+Matching up terms from the beginning of this expansion with those from the end:
+{{begin-eqn}}
+{{eqn | l = 2^n \cos^n \theta
+      | r = e^{n i \theta} + \binom n 1 e^{\paren {n - 2} i \theta} + \binom n 2 e^{\paren {n - 4} i \theta} + \cdots
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \binom n {n - 2} e^{-\paren {n - 4} i \theta} + \binom n {n - 1} e^{-\paren {n - 2} i \theta} + e^{-n i \theta}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2^n} 2 \cos^n \theta
+      | r = \paren {\frac {e^{n i \theta} + e^{-n i \theta} } 2} + \binom n 1 \paren {\frac {e^{\paren {n - 2} i \theta} + e^{-\paren {n - 2} i \theta} } 2}
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \binom n 2 \paren {\frac {e^{\paren {n - 4} i \theta} + e^{-\paren {n - 4} i \theta} } 2} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$
+Now to determine $R_n$.
+The middle term of the sequence $0, 1, \ldots, n$ is $\dfrac n 2$.
+So when $k = \dfrac n 2$ we have $n - 2k = 0$ and $n - k = n - \dfrac n 2 = \dfrac n 2$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \binom n k
+      | r = \frac {n!} {\paren {\paren {n / 2}!} \paren {\paren {n - n / 2}!} }
+      | c = 
+}}
+{{eqn | r = \frac {n!} {\paren {\paren {n / 2}!}^2}
+      | c = 
+}}
+{{end-eqn}}
+So the middle term collapses to:
+{{begin-eqn}}
+{{eqn | l = R_n
+      | r = 2 \frac {n!} {\paren {\paren {n / 2}!}^2} \frac {e^{0 i \theta} } 2
+      | c = 
+}}
+{{eqn | r = \frac {n!} {\paren {\paren {n / 2}!}^2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^{2 n} \theta
+      | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^{2 n}
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \paren {e^{i \theta} + e^{-i \theta} }^{2 n}
+      | c = [[Exponent Combination Laws/Power of Product|Exponent Combination Laws: Power of Product]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \sum^{2 n}_{k \mathop = 0} \binom{2 n} k e^{k i \theta} e^{-\paren {2 n - k} i \theta}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \sum^{2 n}_{k \mathop = 0} \binom{2 n} k e^{-\paren {2 n - 2 k} i \theta}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \paren {\sum^{n \mathop - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \binom {2 n} n e^{\paren {2 n - 2 n} i \theta} + \sum^{2 n}_{k \mathop = n + 1} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} }
+      | c = partitioning the sum
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{2 n}_{k \mathop = n + 1} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} }
+      | c = [[Exponential of Zero]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{n - 1}_{k \mathop = 0} \binom {2 n} {2 n - k} e^{\paren {2 \paren {2 n - k} - 2 n} i \theta} }
+      | c = $k \mapsto 2 n - k$
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{\paren {2 n - 2 k} i \theta} }
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n - 1} } \sum^{n - 1}_{k \mathop = 0} \binom {2 n} k \map \cos {2 n - 2 k} \theta
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Arctangent and Arccotangent}
+Tags: Arctangent Function, Arccotangent Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then:
+: $\arctan x + \operatorname{arccot} x = \dfrac \pi 2$
+where $\arctan$ and $\operatorname{arccot}$ denote [[Definition:Arctangent|arctangent]] and [[Definition:Arccotangent|arccotangent]] respectively.
+\end{theorem}
+
+\begin{proof}
+Let $y \in \R$ such that:
+: $\exists x \in \R: x = \cot \left({y + \dfrac \pi 2}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \cot \left({y + \frac \pi 2}\right)
+      | c = 
+}}
+{{eqn | r = -\tan y
+      | c = [[Cotangent of Angle plus Right Angle]]
+}}
+{{eqn | r = \tan \left({-y}\right)
+      | c = [[Tangent Function is Odd]]
+}}
+{{end-eqn}}
+Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
+Then we can write $-y = \arctan x$.
+But then $\cot \left({y + \dfrac \pi 2}\right) = x$.
+Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.
+Hence $y + \dfrac \pi 2 = \operatorname{arccot} x$.
+That is, $\dfrac \pi 2 = \operatorname{arccot} x + \arctan x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arcsine of Reciprocal equals Arccosecant}
+Tags: Arcsine Function, Arccosecant Function, Reciprocals
+
+\begin{theorem}
+:$\map \arcsin {\dfrac 1 x} = \arccsc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arcsin {\frac 1 x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 x
+      | r = \sin y
+      | c = {{Defof|Arcsine}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \csc y
+      | c = [[Cosecant is Reciprocal of Sine]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arccsc x
+      | r = y
+      | c = {{Defof|Arccosecant}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccosine of Reciprocal equals Arcsecant}
+Tags: Arccosine Function, Arcsecant Function, Reciprocals
+
+\begin{theorem}
+:$\map \arccos {\dfrac 1 x} = \arcsec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arccos {\frac 1 x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 x
+      | r = \cos y
+      | c = {{Defof|Arccosine}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \sec y
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arcsec x
+      | r = y
+      | c = {{Defof|Arcsecant}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arctangent of Reciprocal equals Arccotangent}
+Tags: Arctangent Function, Arccotangent Function, Reciprocals
+
+\begin{theorem}
+:$\map \arctan {\dfrac 1 x} = \arccot x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arctan {\frac 1 x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 x
+      | r = \tan y
+      | c = {{Defof|Arctangent}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \cot y
+      | c = [[Cotangent is Reciprocal of Tangent]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arccot x
+      | r = y
+      | c = {{Defof|Arccotangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Sine is Odd Function}
+Tags: Arcsine Function, Examples of Odd Functions
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arcsin {-x} = -\arcsin x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arcsin {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \sin y:
+      | rr = -\frac \pi 2 \le y \le \frac \pi 2
+      | c = {{Defof|Arcsine}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\sin y:
+      | rr = -\frac \pi 2 \le y \le \frac \pi 2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \sin {-y}:
+      | rr = -\frac \pi 2 \le y \le \frac \pi 2
+      | c = [[Sine Function is Odd]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arcsin x
+      | r = -y
+      | c = {{Defof|Arcsine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccosine of Negative Argument}
+Tags: Arccosine Function
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arccos {-x} = \pi - \arccos x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arccos {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \cos y:
+      | rr= 0 \le y \le \pi
+      | c = {{Defof|Arccosine}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\cos y:
+      | rr= -\pi \le y \le 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \cos {\pi - y}:
+      | rr= 0 \le y \le \pi
+      | c = [[Cosine of Supplementary Angle]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arccos x
+      | r = \pi - y
+      | c = {{Defof|Arccosine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Tangent is Odd Function}
+Tags: Arctangent Function, Examples of Odd Functions
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arctan {-x} = -\arctan x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arctan {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \tan y:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = {{Defof|Arctangent}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\tan y:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \tan {-y}:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = [[Tangent Function is Odd]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arctan x
+      | r = -y
+      | c = {{Defof|Arctangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccotangent of Negative Argument}
+Tags: Arccotangent Function
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arccot {-x} = \pi - \arccot x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arccot {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \cot y:
+      | rr= 0 \le y \le \pi
+      | c = {{Defof|Arccotangent}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\cot y:
+      | rr= -\pi \le y \le 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \cot {\pi - y}:
+      | rr= 0 \le y \le \pi
+      | c = [[Cotangent of Supplementary Angle]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arccot x
+      | r = \pi - y
+      | c = {{Defof|Arccotangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arcsecant of Negative Argument}
+Tags: Arcsecant Function
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arcsec {-x} = \pi - \arcsec x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arcsec {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \sec y:
+      | rr= 0 \le y \le \pi
+      | c = {{Defof|Arcsecant}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\sec y:
+      | rr= -\pi \le y \le 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \sec {\pi - y}:
+      | rr= 0 \le y \le \pi
+      | c = [[Cosecant of Supplementary Angle]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arcsec x
+      | r = \pi - y
+      | c = {{Defof|Arcsecant}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Cosecant is Odd Function}
+Tags: Arccosecant Function, Examples of Odd Functions
+
+\begin{theorem}
+Everywhere that the function is defined:
+:$\map \arccsc {-x} = -\arccsc x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \arccsc {-x}
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x
+      | r = \csc y:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = {{Defof|Arccosecant}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = -\csc y:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \map \csc {-y}:
+      | rr= -\frac \pi 2 \le y \le \frac \pi 2
+      | c = [[Cosecant Function is Odd]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \arccsc x
+      | r = -y
+      | c = {{Defof|Arccosecant}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Tangents}
+Tags: Triangles, Tangent Function, Named Theorems
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
+Then:
+:$\dfrac {a + b} {a - b} = \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }$
+\end{theorem}
+
+\begin{proof}
+Let $d = \dfrac a {\sin A}$.
+From the [[Law of Sines]], let:
+:$d = \dfrac a {\sin A} = \dfrac b {\sin B}$
+so that:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = d \sin A
+      | c = 
+}}
+{{eqn | l = b
+      | r = d \sin B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {a + b} {a - b}
+      | r = \frac {d \sin A + d \sin B} {d \sin A - d \sin B}
+      | c = 
+}}
+{{eqn | r = \frac {\sin A + \sin B} {\sin A - \sin B}
+      | c = 
+}}
+{{eqn-intertext|Let $C {{=}} \frac 1 2 \paren {A + B}$ and $D {{=}} \frac 1 2 \paren {A - B}$, and proceed as follows:}}
+{{eqn | ll= \leadsto
+      | l = \frac {a + b} {a - b}
+      | r = \frac {2 \sin C \cos D} {\sin A - \sin B}
+      | c = [[Prosthaphaeresis Formula for Sine plus Sine]]
+}}
+{{eqn | r = \frac {2 \sin C \cos D} {2 \sin D \cos C}
+      | c = [[Prosthaphaeresis Formula for Sine minus Sine]]
+}}
+{{eqn | r = \frac {\frac {\sin C} {\cos C} } {\frac {\sin D} {\cos D} }
+      | c = dividing top and bottom by $\cos C \cos D$
+}}
+{{eqn | r = \frac {\tan C} {\tan D}
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }
+      | c = substituting back for $C$ and $D$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Angle of Triangle by Semiperimeter}
+Tags: Triangles, Sine Function
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
+Then:
+: $\sin A = \dfrac 2 {b c} \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+where $\sin$ denotes [[Definition:Sine|sine]] and $s$ is the [[Definition:Semiperimeter|semiperimeter]]: $s = \dfrac {a + b + c} 2$.
+\end{theorem}
+
+\begin{proof}
+Let $Q$ be the [[Definition:Area|area]] of $\triangle ABC$.
+From [[Area of Triangle in Terms of Two Sides and Angle]]:
+:$Q = \dfrac {b c \sin A} 2$
+From [[Heron's Formula]]:
+:$Q = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+Equating the two:
+:$\dfrac {b c \sin A} 2 = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+from which follows the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/Negation, Conjunction, Disjunction and Implication}
+Tags: Functional Completeness
+
+\begin{theorem}
+: $\left\{{\neg, \land, \lor, \implies}\right\}$: [[Definition:Logical Not|Not]], [[Definition:Conjunction|And]], [[Definition:Disjunction|Or]] and [[Definition:Conditional|Implies]]
+\end{theorem}
+
+\begin{proof}
+From [[Functional Completeness over Finite Number of Arguments]], it suffices to consider binary [[Definition:Truth Function|truth functions]].
+From [[Count of Truth Functions]], there are $16$ of these.
+These are enumerated in [[Binary Truth Functions]], and are analysed in turn as follows.
+=== Constant Functions ===
+There are two [[Definition:Constant Mapping|constant functions]]:
+{{begin-eqn}}
+{{eqn | l = f_F \left({p, q}\right)
+      | r = F
+}}
+{{eqn | l = f_T \left({p, q}\right)
+      | r = T
+}}
+{{end-eqn}}
+From [[Biconditional Properties]] and [[Exclusive Or Properties]]:
+{{begin-eqn}}
+{{eqn | l = p \iff p
+      | o = \dashv \vdash
+      | r = \top \dashv \vdash f_T \left({p, q}\right)
+}}
+{{eqn | l = p \oplus p
+      | o = \dashv \vdash
+      | r = \bot \dashv \vdash f_F \left({p, q}\right)
+}}
+{{end-eqn}}
+where $\oplus$ and $\iff$ denote [[Definition:Exclusive Or|exclusive or]] and [[Definition:Biconditional|biconditional]] respectively.
+So both of the constant functions can be expressed in terms of $\oplus$ and $\iff$.
+{{qed|lemma}}
+=== Equivalence and Non-Equivalence ===
+By definition of [[Definition:Exclusive Or|exclusive or]]:
+: $p \oplus q \dashv \vdash \neg \left({p \iff q}\right)$.
+Thus $\oplus$ can be expressed in terms of $\neg$ and $\iff$.
+By definition of [[Definition:Biconditional|biconditional]]:
+: $p \iff q \dashv \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$
+Thus $\iff$ can be expressed in terms of $\implies$ and $\land$.
+{{qed|lemma}}
+=== Projections and Negated Projections ===
+There are two [[Definition:Projection (Mapping Theory)|projections]]:
+: $\operatorname{pr}_1 \left({p, q}\right) = p$
+: $\operatorname{pr}_2 \left({p, q}\right) = q$
+We note that:
+: $p \land p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$
+: $p \lor p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$
+and similarly for $\operatorname{pr}_2 \left({p, q}\right)$.
+So the [[Definition:Projection (Mapping Theory)|projections]] can be expressed in terms of either $\land$ or $\lor$.
+There are two [[Definition:Logical Not|negated]] [[Definition:Projection (Mapping Theory)|projections]]:
+: $\overline {\operatorname{pr}_1} \left({p, q}\right) = \neg p$
+: $\overline {\operatorname{pr}_2} \left({p, q}\right) = \neg q$
+It immediately follows from the above that these can be expressed in terms of either:
+: $\neg$ and $\land$
+or:
+: $\neg$ and $\lor$
+{{qed|lemma}}
+=== NAND and NOR ===
+There are the [[Definition:Logical NAND|NAND]] and [[Definition:Logical NOR|NOR]] operators:
+: $p \uparrow q$
+: $p \downarrow q$
+By definition of [[Definition:Logical NAND|NAND]]:
+: $p \uparrow q \dashv \vdash \neg \left({p \land q}\right)$
+By definition of [[Definition:Logical NOR|NOR]]:
+: $p \downarrow q \dashv \vdash \neg \left({p \lor q}\right)$
+So:
+: $\uparrow$ can be expressed in terms of $\neg$ and $\land$
+: $\downarrow$ can be expressed in terms of $\neg$ and $\lor$
+{{qed|lemma}}
+=== Conjunction, Disjunction, Conditional ===
+There are the [[Definition:Conjunction|conjunction]] and [[Definition:Disjunction|disjunction]] operators:
+: $p \land q$
+: $p \lor q$
+There are two [[Definition:Conditional|conditionals]]:
+: $p \implies q$
+: $q \implies p$
+There are two  [[Definition:Logical Not|negated]] [[Definition:Conditional|conditionals]]:
+: $\neg \left({p \implies q}\right)$
+: $\neg \left({q \implies p}\right)$
+All of the above are already expressed in terms of $\neg, \land, \lor, \implies$.
+{{qed|lemma}}
+Thus all sixteen of the [[Binary Truth Functions]] can be expressed in terms of:
+: $\neg, \land, \lor, \implies$
+That is:
+: $\left\{{\neg, \land, \lor, \implies}\right\}$
+is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+[[Category:Functional Completeness]]
+0qv4fsb1giyp2qwmxxefz3707f3rl8b
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/Negation and Disjunction}
+Tags: Functional Completeness
+
+\begin{theorem}
+:$\set {\neg, \lor}$: [[Definition:Logical Not|Not]] and [[Definition:Disjunction|Or]]
+\end{theorem}
+
+\begin{proof}
+From [[Functionally Complete Logical Connectives/Negation and Conjunction|Functionally Complete Logical Connectives: Negation and Conjunction]], $\set {\neg, \land}$ is [[Definition:Functionally Complete|functionally complete]].
+That is: any expression can be expressed in terms of $\neg$ and $\land$.
+From [[De Morgan's Laws (Logic)/Conjunction|De Morgan's laws: Conjunction]], we have that:
+:$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$
+Thus all occurrences of $\land$ can be replaced by $\lor$ and $\neg$.
+Thus any expression can be expressed in terms of $\neg$ and $\lor$.
+That is: $\set {\neg, \lor}$ is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/Negation and Conjunction}
+Tags: Functional Completeness
+
+\begin{theorem}
+:$\set {\neg, \land}$: [[Definition:Logical Not|Not]] and [[Definition:Conjunction|And]]
+\end{theorem}
+
+\begin{proof}
+From [[Functionally Complete Logical Connectives/Negation, Conjunction, Disjunction and Implication|Functionally Complete Logical Connectives: Negation, Conjunction, Disjunction and Implication]], all sixteen of the [[Binary Truth Functions|binary truth functions]] can be expressed in terms of $\neg, \land, \lor, \implies$.
+From [[Conjunction and Implication]], we have that:
+:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
+From [[De Morgan's Laws (Logic)/Disjunction|De Morgan's laws: Disjunction]], we have that:
+:$p \lor q \dashv \vdash \neg \paren {\neg p \land \neg q}$
+So any instance of either $\implies$ or $\lor$ can be replaced identically with one using just $\neg$ and $\land$.
+It follows that $\set {\neg, \land}$ is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/Negation and Conditional}
+Tags: Functional Completeness
+
+\begin{theorem}
+:$\set {\neg, \implies}$: [[Definition:Logical Not|Not]] and [[Definition:Conditional|Implies]]
+\end{theorem}
+
+\begin{proof}
+From [[Functionally Complete Logical Connectives/Negation and Conjunction|Functionally Complete Logical Connectives: Negation and Conjunction]], we can represent any boolean expression in terms of $\land$ and $\neg$.
+From [[Conjunction and Implication]], we have that:
+:$p \land q \dashv \vdash \neg \paren {p \implies \neg q}$
+So it follows that we can replace all occurrences of $\land$ by $\implies$ and $\neg$.
+Thus $\set {\neg, \implies}$ is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/Conjunction, Negation and Disjunction}
+Tags: Functional Completeness
+
+\begin{theorem}
+:$\set {\neg, \land, \lor}$: [[Definition:Logical Not|Not]], [[Definition:Conjunction|And]] and [[Definition:Disjunction|Or]]
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Stronger Statement|stronger]] results:
+:[[Functionally Complete Logical Connectives/Negation and Disjunction|Functionally Complete Logical Connectives: Negation and Disjunction]]:
+::the [[Definition:Set|set]] of [[Definition:Logical Connective|logical connectives]]: $\set {\neg, \lor}$ is [[Definition:Functionally Complete|functionally complete]]
+:[[Functionally Complete Logical Connectives/Negation and Conjunction|Functionally Complete Logical Connectives: Negation and Conjunction]]:
+::the [[Definition:Set|set]] of [[Definition:Logical Connective|logical connectives]]: $\set {\neg, \land}$ is [[Definition:Functionally Complete|functionally complete]]
+it follows directly that $\set {\neg, \land, \lor}$ is likewise [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+[[Category:Functional Completeness]]
+eh8et9jrx7lv8b1gvg1ipjzycdrvr03
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/NAND}
+Tags: Functional Completeness, Logical NAND
+
+\begin{theorem}
+:$\set {\uparrow}$: [[Definition:Logical NAND|NAND]]
+\end{theorem}
+
+\begin{proof}
+From [[Functionally Complete Logical Connectives/Negation and Conjunction|Functionally Complete Logical Connectives: Negation and Conjunction]], any boolean expression can be expressed in terms of $\land$ and $\neg$.
+From [[NAND with Equal Arguments]]:
+:$\neg p \dashv \vdash p \uparrow p$
+From [[Conjunction in terms of NAND]]:
+:$p \land q \dashv \vdash \paren {p \uparrow q} \uparrow \paren {p \uparrow q}$
+demonstrating that $p \land q$ is expressed solely in terms of $\uparrow$.
+Thus any boolean expression can be represented solely in terms of $\uparrow$.
+That is, $\set {\uparrow}$ is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Logical Connectives/NOR}
+Tags: Functional Completeness, Logical NOR
+
+\begin{theorem}
+:$\set {\downarrow}$: [[Definition:Logical NOR|NOR]]
+\end{theorem}
+
+\begin{proof}
+From [[Functionally Complete Logical Connectives/Negation and Disjunction|Functionally Complete Logical Connectives: Negation and Disjunction]], any boolean expression can be expressed in terms of $\lor$ and $\neg$.
+From [[NOR with Equal Arguments]]:
+:$\neg p \dashv \vdash p \downarrow p$
+From [[Disjunction in terms of NOR]]:
+:$p \lor q \dashv \vdash \paren {p \downarrow q} \downarrow \paren {p \downarrow q}$
+demonstrating that $p \lor q$ can be represented solely in terms of $\downarrow$.
+That is, $\set {\downarrow}$ is [[Definition:Functionally Complete|functionally complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functionally Complete Singleton Sets}
+Tags: Functional Completeness
+
+\begin{theorem}
+The only [[Definition:Binary Logical Connective|binary logical connectives]] that form [[Definition:Singleton|singleton sets]] which are [[Definition:Functionally Complete|functionally complete]] are [[Definition:Logical NAND|NAND]] and [[Definition:Logical NOR|NOR]].
+\end{theorem}
+
+\begin{proof}
+Let $\uparrow$ and $\downarrow$ denote [[Definition:Logical NAND|NAND]] and [[Definition:Logical NOR|NOR]] respectively.
+From:
+: [[NAND is Functionally Complete]]
+and
+: [[NOR is Functionally Complete]]
+the [[Definition:Singleton|singleton sets]] $\left\{{\uparrow}\right\}$ and $\left\{{\downarrow}\right\}$ are [[Definition:Functionally Complete|functionally complete]].
+Suppose $\circ$ is a [[Definition:Binary Logical Connective|binary logical connective]] such that $\left\{{\circ}\right\}$ is a [[Definition:Functionally Complete|functionally complete set]].
+The [[Definition:Unary Logical Connective|unary logical connective]] $\neg$ has to be equivalent to some formula:
+: $\cdots \circ \left({p \circ p}\right) \circ \cdots$
+Suppose some [[Definition:Boolean Interpretation|interpretation]] $v$ assigns $T$ to $p$.
+Then $v \left({\neg p}\right) = F$
+So:
+: $v \left({\cdots \circ \left({p \circ p}\right) \circ \cdots}\right) = F$
+So it has to be true that $T \circ T = F$, otherwise:
+: $v \left({\cdots \circ \left({T \circ T}\right) \circ \cdots}\right) = v \left({\cdots \circ \left({T}\right) \circ \cdots}\right) = T$
+Similarly:
+:$F \circ F = T$
+Now consider $T \circ F$ and $F \circ T$.
+Suppose $T \circ F = F$.
+If $F \circ T = T$, then we have the function defined by this [[Definition:Truth Table|truth table]]:
+{| border="1" cellpadding="4" cellspacing="1" style="text-align:center;"
+|- 
+! $v \left({p}\right)$
+! $v \left({q}\right)$
+! $v \left({p \circ q}\right)$
+|-
+| $F$ || $F$ || $T$
+|-
+| $F$ || $T$ || $T$
+|-
+| $T$ || $F$ || $F$
+|-
+| $T$ || $T$ || $F$
+|}
+which is $\neg p$, and hence only negation would be definable.
+So if $T \circ F = F$ we need $F \circ T = F$.
+This gives the [[Definition:Truth Table|truth table]] for the [[Definition:Logical NOR|NOR]] function:
+{| border="1" cellpadding="4" cellspacing="1" style="text-align:center;"
+|- 
+! $v \left({p}\right)$
+! $v \left({q}\right)$
+! $v \left({p \circ q}\right)$
+|-
+| $F$ || $F$ || $T$
+|-
+| $F$ || $T$ || $F$
+|-
+| $T$ || $F$ || $F$
+|-
+| $T$ || $T$ || $F$
+|}
+... which we have seen is [[Definition:Functionally Complete|functionally complete]].
+Similarly, Suppose $T \circ F = T$.
+If $F \circ T = F$, then we have the function defined by this [[Definition:Truth Table|truth table]]:
+{| border="1" cellpadding="4" cellspacing="1" style="text-align:center;"
+|- 
+! $v \left({p}\right)$
+! $v \left({q}\right)$
+! $v \left({p \circ q}\right)$
+|-
+| $F$ || $F$ || $T$
+|-
+| $F$ || $T$ || $F$
+|-
+| $T$ || $F$ || $T$
+|-
+| $T$ || $T$ || $F$
+|}
+which is $\neg q$, and hence only negation would be definable.
+So if $T \circ F = T$ we need $F \circ T = T$.
+This gives the [[Definition:Truth Table|truth table]] for the [[Definition:Logical NAND|NAND]] function:
+{| border="1" cellpadding="4" cellspacing="1" style="text-align:center;"
+|- 
+! $v \left({p}\right)$
+! $v \left({q}\right)$
+! $v \left({p \circ q}\right)$
+|-
+| $F$ || $F$ || $T$
+|-
+| $F$ || $T$ || $T$
+|-
+| $T$ || $F$ || $T$
+|-
+| $T$ || $T$ || $F$
+|}
+... which we have seen is [[Definition:Functionally Complete|functionally complete]].
+Thus it follows that there can be no [[Definition:Functionally Complete|functionally complete]] [[Definition:Binary Logical Connective|binary logical connectives]] apart from [[Definition:Logical NAND|NAND]] and [[Definition:Logical NOR|NOR]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR with Equal Arguments/Proof 1}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then for any [[Definition:Proposition|proposition]] $p$:
+:$p \downarrow p \dashv \vdash \neg p$
+That is, the [[Definition:Logical NOR|NOR]] of a proposition with itself corresponds to the [[Definition:Negation|negation]] operator.
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \downarrow p \vdash \neg p}}
+{{Premise|1|p \downarrow p}}
+{{SequentIntro|2|1|\neg \left({p \lor p}\right)|1|Definition of [[Definition:Logical NOR|Logical NOR]]}}
+{{Idempotence|3|1|\neg p|2|Disjunction}}
+{{EndTableau}}
+{{BeginTableau|\neg p \vdash p \downarrow p}}
+{{Premise|1|\neg p}}
+{{Idempotence|2|1|\neg \left({p \lor p}\right)|1|Disjunction}}
+{{SequentIntro|3|1|p \downarrow p|2|Definition of [[Definition:Logical NOR|Logical NOR]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR with Equal Arguments/Proof 2}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then for any [[Definition:Proposition|proposition]] $p$:
+:$p \downarrow p \dashv \vdash \neg p$
+That is, the [[Definition:Logical NOR|NOR]] of a proposition with itself corresponds to the [[Definition:Negation|negation]] operator.
+\end{theorem}
+
+\begin{proof}
+Apply the [[Method of Truth Tables]]:
+:$\begin{array}{|ccc||cc|} \hline
+p & \downarrow & p & \neg & p \\
+\hline
+F & T & F & T & F \\
+T & F & T & F & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NAND with Equal Arguments/Proof 1}
+Tags: Logical NAND
+
+\begin{theorem}
+Let $\uparrow$ signify the [[Definition:Logical NAND|NAND]] operation.
+Then, for any [[Definition:Proposition|proposition]] $p$:
+:$p \uparrow p \dashv \vdash \neg p$
+That is, the [[Definition:Logical NAND|NAND]] of a proposition with itself corresponds to the [[Definition:Negation|negation]] operation.
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \uparrow p \vdash \neg p}}
+{{Premise|1|p \uparrow p}}
+{{SequentIntro|2|1|\neg \left({p \land p}\right)|1|Definition of [[Definition:Logical NAND|Logical NAND]]}}
+{{Idempotence|3|1|\neg p|2|Conjunction}}
+{{EndTableau}}
+{{BeginTableau|\neg p \vdash p \uparrow p}}
+{{Premise|1|\neg p}}
+{{Idempotence|2|1|\neg \left({p \land p}\right)|1|Conjunction}}
+{{SequentIntro|3|1|p \uparrow p|2|Definition of [[Definition:Logical NAND|Logical NAND]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NAND with Equal Arguments/Proof 2}
+Tags: Logical NAND, Truth Table Proofs
+
+\begin{theorem}
+Let $\uparrow$ signify the [[Definition:Logical NAND|NAND]] operation.
+Then, for any [[Definition:Proposition|proposition]] $p$:
+:$p \uparrow p \dashv \vdash \neg p$
+That is, the [[Definition:Logical NAND|NAND]] of a proposition with itself corresponds to the [[Definition:Negation|negation]] operation.
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]]:
+:$\begin{array}{|ccc||cc|} \hline
+p & \uparrow & p & \neg & p \\
+\hline
+F & T & F & T & F \\
+T & F & T & F & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conditional and Converse are not Equivalent}
+Tags: Implication
+
+\begin{theorem}
+A [[Definition:Conditional|conditional statement]]:
+:$p \implies q$
+is not [[Definition:Logical Equivalence|logically equivalent]] to its [[Definition:Converse Statement|converse]]:
+:$q \implies p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition:
+:$\paren {p \implies q} \iff \paren {q \implies p}$
+$\begin{array}{|ccc|c|ccc|} \hline
+p & \implies & q) & \iff & (q & \implies & p) \\
+\hline
+F & T & F & T & F & T & F \\
+F & T & T & F & T & F & F \\
+T & F & F & F & F & T & T \\
+T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] do not match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR is Commutative/Proof 1}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then, for any two [[Definition:Proposition|propositions]] $p$ and $q$:
+:$p \downarrow q \dashv \vdash q \downarrow p$
+That is, [[Definition:Logical NOR|NOR]] is [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \downarrow q \vdash q \downarrow p}}
+{{Premise|1|p \downarrow q}}
+{{SequentIntro|2|1|\neg \paren {p \lor q}|1|{{Defof|Logical NOR}} }}
+{{Commutation|3|1|\neg \paren {q \lor p}|2|Disjunction}}
+{{SequentIntro|4|1|q \uparrow p|3|{{Defof|Logical NOR}} }}
+{{EndTableau}}
+{{qed|lemma}}
+{{BeginTableau|q \downarrow p \vdash p \uparrow q}}
+{{Premise|1|q \downarrow p}}
+{{SequentIntro|2|1|\neg \paren {q \lor p}|1|{{Defof|Logical NOR}} }}
+{{Commutation|3|1|\neg \paren {p \lor q}|2|Disjunction}}
+{{SequentIntro|4|1|p \downarrow q|3|{{Defof|Logical NOR}} }}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR is Commutative/Proof 2}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then, for any two [[Definition:Proposition|propositions]] $p$ and $q$:
+:$p \downarrow q \dashv \vdash q \downarrow p$
+That is, [[Definition:Logical NOR|NOR]] is [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+Apply the [[Method of Truth Tables]]:
+:$\begin{array}{|ccc||ccc|} \hline
+p & \downarrow & q & q & \downarrow & p \\
+\hline
+F & T & F & F & T & F \\
+F & F & T & T & F & F \\
+T & F & F & F & F & T \\
+T & F & T & T & F & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR is not Associative/Proof 1}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then there exist [[Definition:Proposition|propositions]] $p,q,r$ such that:
+:$p \downarrow \left({q \downarrow r}\right) \not \vdash \left({p \downarrow q}\right) \downarrow r$
+That is, [[Definition:Logical NOR|NOR]] is not [[Definition:Associative|associative]].
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({ p \downarrow \left({q \downarrow r}\right) \implies \left({p \downarrow q}\right) \downarrow r }\right)}}
+{{Assumption|1|\neg p \land r}}
+{{Simplification|2|1|\neg p|1|1}}
+{{Simplification|3|1|r|1|2}}
+{{Addition|4|1|q \lor r|3|2}}
+{{DoubleNegIntro|5|1|\neg \neg \left({q \lor r}\right)|4}}
+{{SequentIntro|6|1|\neg \left({q \downarrow r}\right)|5|Definition of [[Definition:Logical NOR|Logical NOR]]}}
+{{Conjunction|7|1|\neg p \land \neg \left({q \downarrow r}\right)|2|6}}
+{{DeMorgan|8|1|\neg \left({p \lor \left({q \downarrow r}\right) }\right)|7|Conjunction of Negations}}
+{{SequentIntro|9|1|p \downarrow \left({q \downarrow r}\right)|8|Definition of [[Definition:Logical NOR|Logical NOR]]}}
+{{Addition|10|1|\left({p \downarrow q}\right) \lor r|3|2}}
+{{DoubleNegIntro|11|1|\neg \neg \left({\left({p \downarrow q}\right) \lor r}\right)|10}}
+{{SequentIntro|12|1|\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)|11|Definition of [[Definition:Logical NOR|Logical NOR]]}}
+{{DoubleNegIntro|13|1|\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)|9}}
+{{Conjunction|14|1|\left({\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)}\right) \land \left({\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)}\right)|13|12}}
+{{DeMorgan|15|1|\neg \left({\neg \left({p \downarrow \left({q \downarrow r}\right)}\right) \lor \left({\left({p \downarrow q}\right) \downarrow r}\right)}\right)|14|Conjunction of Negations}}
+{{SequentIntro|16|1
+  |\neg \left({ p \downarrow \left({q \downarrow r}\right) \implies \left({p \downarrow q}\right) \downarrow r }\right)
+  |15
+  |[[Rule of Material Implication]]
+}}
+{{EndTableau}}
+Taking $p = \bot$ and $r = \top$, we have $\vdash \neg p \land r$, discharging the last assumption.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{NOR is not Associative/Proof 2}
+Tags: Logical NOR
+
+\begin{theorem}
+Let $\downarrow$ signify the [[Definition:Logical NOR|NOR]] operation.
+Then there exist [[Definition:Proposition|propositions]] $p, q, r$ such that:
+:$p \downarrow \paren {q \downarrow r} \not \vdash \paren {p \downarrow q} \downarrow r$
+That is, [[Definition:Logical NOR|NOR]] is not [[Definition:Associative|associative]].
+\end{theorem}
+
+\begin{proof}
+Apply the [[Method of Truth Tables]]:
+:$\begin{array}{|ccccc||ccccc|} \hline
+p & \downarrow & (q & \downarrow & r) & (p & \downarrow & q) & \downarrow & r \\
+\hline
+F & F & F & T & F & F & T & F & F & F \\
+F & T & F & F & T & F & T & F & F & T \\
+F & T & T & F & F & F & F & T & T & F \\
+F & T & T & F & T & F & F & T & F & T \\
+T & F & F & T & F & T & F & F & T & F \\
+T & F & F & F & T & T & F & F & F & T \\
+T & F & T & F & F & T & F & T & T & F \\
+T & F & T & F & T & T & F & T & F & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] do not match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Predicate Logic)/Assertion of Universality}
+Tags: Universal Quantifier, Existential Quantifier, De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\forall x: \map P x \dashv \vdash \neg \exists x: \neg \map P x$
+::''If everything '''is''', there exists nothing that '''is not'''.''
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\forall x: \map P x \vdash \neg \exists x: \neg \map P x}}
+{{Premise|1|\forall x: \map P x}}
+{{Assumption|2|\exists x: \neg \map P x}}
+{{TableauLine|n = 3|pool = 2|f = \neg \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 2|c = for some arbitrary $\mathbf a$}}
+{{TableauLine|n = 4|pool = 1|f = \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 1}}
+{{NonContradiction|5|1, 2|3|4}}
+{{Contradiction|6|1|\neg \exists x: \neg \map P x|2|5}}
+{{EndTableau|lemma}}
+{{BeginTableau|\neg \exists x: \neg \map P x \vdash \forall x: \map P x}}
+{{Premise|1|\neg \exists x: \neg \map P x}}
+{{Assumption|2|\neg \forall x: \map P x}}
+{{SequentIntro|3|2|\exists x: \neg \map P x|2|[[Denial of Universality]]}}
+{{NonContradiction|4|1, 2|1|3}}
+{{Reductio|5|1|\forall x: \map P x|2|4}}
+{{EndTableau|qed}}
+{{LEM|Reductio ad Absurdum}}
+{{Namedfor|Augustus De Morgan|cat = De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Predicate Logic)/Denial of Existence}
+Tags: Universal Quantifier, Existential Quantifier, De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\forall x: \neg \map P x \dashv \vdash \neg \exists x: \map P x$
+::''If everything '''is not''', there exists nothing that '''is'''.''
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\forall x: \neg \map P x \vdash \neg \exists x: \map P x}}
+{{Premise|1|\forall x: \neg \map P x}}
+{{Assumption|2|\exists x: \map P x}}
+{{TableauLine|n = 3|pool = 2|f = \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 2|c = for an arbitrary $\mathbf a$}}
+{{TableauLine|n = 4|pool = 1|f = \neg \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 3}}
+{{NonContradiction|5|1, 2|3|4}}
+{{Contradiction|6|1|\neg \exists x: \map P x|2|5}}
+{{EndTableau|lemma}}
+{{BeginTableau|\neg \exists x: \map P x \vdash \forall x: \neg \map P x}}
+{{Premise|1|\neg \exists x: \map P x}}
+{{Assumption|2|\map P {\mathbf a}|for some arbitrary $\mathbf a$}}
+{{TableauLine|n = 3|pool = 2|f = \exists x: \map P x|rlnk = Existential Generalisation|rtxt = Existential Generalisation|dep = 2}}
+{{NonContradiction|4|1, 2|1|3}}
+{{Contradiction|5|1|\neg \map P {\mathbf a}|2|4}}
+{{TableauLine|n = 6|pool = 1|f = \forall x: \neg \map P x|rlnk = Universal Generalisation|rtxt = Universal Generalisation|dep = 5|c = as $\mathbf a$ was arbitrary}}
+{{EndTableau|qed}}
+{{Namedfor|Augustus De Morgan|cat = De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Predicate Logic)/Denial of Universality}
+Tags: Universal Quantifier, Existential Quantifier, De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg \forall x: \map P x \dashv \vdash \exists x: \neg \map P x$
+::''If not everything '''is''', there exists something that '''is not'''.''
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \forall x: \map P x \vdash \exists x: \neg \map P x}}
+{{Premise|1|\neg \forall x: \map P x}}
+{{Assumption|2|\neg \exists x: \neg \map P x}}
+{{Assumption|3|\neg \map P {\mathbf a}|for an arbitrary $\mathbf a$}}
+{{TableauLine|n = 4|pool = 3|f = \exists x: \neg \map P x|rlnk = Existential Generalisation|rtxt = Existential Generalisation|dep = 3}}
+{{NonContradiction|5|2, 3|2|4}}
+{{Reductio|6|2|\map P {\mathbf a}|3|5}}
+{{TableauLine|n = 7|pool = 1, 2|f = \forall x: \map P x|rlnk = Universal Generalisation|rtxt = Universal Generalisation|dep = 6|c = as $\mathbf a$ was arbitrary}}
+{{NonContradiction|8|2|1|7}}
+{{Reductio|9|1|\exists x: \neg \map P x|2|8}}
+{{EndTableau|lemma}}
+{{BeginTableau|\exists x: \neg \map P x \vdash \neg \forall x: \map P x}}
+{{Premise|1|\exists x: \neg \map P x}}
+{{Assumption|2|\forall x: \map P x}}
+{{TableauLine|n = 3|pool = 1|f = \neg \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 1}}
+{{TableauLine|n = 4|pool = 2|f = \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 2}}
+{{NonContradiction|5|1, 2|3|4}}
+{{Contradiction|6|1|\neg \forall x: \map P x|2|5}}
+{{EndTableau|qed}}
+{{LEM|Reductio ad Absurdum}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Predicate Logic)/Assertion of Existence}
+Tags: Universal Quantifier, Existential Quantifier, De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg \forall x: \neg \map P x \dashv \vdash \exists x: \map P x$
+::''If not everything '''is not''', there exists something that '''is'''.''
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \forall x: \neg \map P x \vdash \exists x: \map P x}}
+{{Premise|1|\neg \forall x: \neg \map P x}}
+{{Assumption|2|\neg \exists x: \map P x}}
+{{SequentIntro|3|2|\forall x: \neg \map P x|2|[[Denial of Existence]]}}
+{{NonContradiction|4|1, 2|1|3}}
+{{Reductio|5|1|\exists x: \map P x|2|4}}
+{{EndTableau|lemma}}
+{{BeginTableau|\exists x: \map P x \vdash \neg \forall x: \neg \map P x}}
+{{Premise|1|\exists x: \map P x}}
+{{Assumption|2|\forall x: \neg \map P x}}
+{{TableauLine|n = 3|pool = 1|f = \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 1|for some arbitrary $\mathbf a$}}
+{{TableauLine|n = 4|pool = 2|f = \neg \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 2}}
+{{NonContradiction|5|1, 2|3|4}}
+{{Contradiction|6|1|\neg \forall x: \neg \map P x|2|5}}
+{{EndTableau|qed}}
+{{LEM|Reductio ad Absurdum}}
+{{Namedfor|Augustus De Morgan|cat = De Morgan}}
+\end{proof}<|endoftext|>
+\section{Conjunction in terms of NAND}
+Tags: Logical NAND, Conjunction
+
+\begin{theorem}
+:$p \land q \dashv \vdash \paren {p \uparrow q} \uparrow \paren {p \uparrow q}$
+where $\land$ denotes [[Definition:Conjunction|logical conjunction]] and $\uparrow$ denotes [[Definition:Logical NAND|logical NAND]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \land q
+      | o = \dashv \vdash
+      | r = \neg \neg \paren {p \land q}
+      | c = [[Double Negation]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {p \uparrow q}
+      | c = {{Defof|Logical NAND}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \paren {p \uparrow q} \uparrow \paren {p \uparrow q}
+      | c = [[NAND with Equal Arguments]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Disjunction in terms of NOR}
+Tags: Logical NOR, Disjunction
+
+\begin{theorem}
+:$p \lor q \dashv \vdash \paren {p \downarrow q} \downarrow \paren {p \downarrow q}$
+where $\lor$ denotes [[Definition:Disjunction|logical disjunction]] and $\downarrow$ denotes [[Definition:Logical NOR|logical NOR]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \lor q
+      | o = \dashv \vdash
+      | r = \neg \neg \paren {p \lor q}
+      | c = [[Double Negation]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {p \downarrow q}
+      | c = {{Defof|Logical NOR}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \paren {p \downarrow q} \downarrow \paren {p \downarrow q}
+      | c = [[NOR with Equal Arguments]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Logical NOR]]
+[[Category:Disjunction]]
+4dw5e72qt5juo7ko2pxvsrci6p0nyyc
+\end{proof}<|endoftext|>
+\section{Disjunction in terms of NAND}
+Tags: Logical NAND, Disjunction
+
+\begin{theorem}
+:$p \lor q \dashv \vdash \paren {p \uparrow p} \uparrow \paren {q \uparrow q}$
+where $\lor$ denotes [[Definition:Disjunction|logical disjunction]] and $\uparrow$ denotes [[Definition:Logical NAND|logical NAND]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \lor q
+      | o = \dashv \vdash
+      | r = \neg \paren {\neg p \land \neg q}
+      | c = [[De Morgan's Laws (Logic)/Disjunction|De Morgan's Laws (Logic): Disjunction]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg p \uparrow \neg q
+      | c = {{Defof|Logical NAND}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \paren {p \uparrow p} \uparrow \paren {q \uparrow q}
+      | c = [[NAND with Equal Arguments]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication in terms of NAND}
+Tags: Logical NAND, Implication, Implication in terms of NAND
+
+\begin{theorem}
+:$p \implies q \dashv \vdash p \uparrow \paren {q \uparrow q}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \implies q
+      | o = \dashv \vdash
+      | r = \neg \paren {p \land \neg q}
+      | c = [[Implication Equivalent to Negation of Conjunction with Negative]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = p \uparrow \neg q
+      | c = Definition of [[Definition:Logical NAND|NAND]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = p \uparrow \paren {q \uparrow q}
+      | c = [[NAND with Equal Arguments]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \implies q
+      | o = \dashv \vdash
+      | r = \neg p \lor q
+      | c = [[Rule of Material Implication]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg p \lor \neg \neg q
+      | c = [[Double Negation Introduction]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = p \uparrow \neg q
+      | c = [[NAND as Disjunction of Negations]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = p \uparrow \left({q \uparrow q}\right)
+      | c = [[NAND with Equal Arguments]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional in terms of NAND}
+Tags: Logical NAND, Biconditional
+
+\begin{theorem}
+:$p \iff q \dashv \vdash \paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \uparrow \paren {p \uparrow q}$
+where $\iff$ denotes [[Definition:Biconditional|logical biconditional]] and $\uparrow$ denotes [[Definition:Logical NAND|logical NAND]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \iff q
+      | o = \dashv \vdash
+      | r = \neg \paren {p \oplus q}
+      | c = [[Exclusive Or is Negation of Biconditional]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {\paren {p \lor q} \land \neg \paren {p \land q} }
+      | c = {{Defof|Exclusive Or}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {\paren {p \lor q} \land \paren {p \uparrow q} }
+      | c = {{Defof|Logical NAND}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {\paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \land \paren {p \uparrow q} }
+      | c = [[Disjunction in terms of NAND]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \uparrow \paren {p \uparrow q}
+      | c = {{Defof|Logical NAND}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping is Injection and Surjection iff Inverse is Mapping}
+Tags: Injections, Surjections, Inverse Mappings, Equivalence of Definitions of Bijection, Mapping is Injection and Surjection iff Inverse is Mapping
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+: $f: S \to T$ can be defined as a [[Definition:Bijection/Definition 1|bijection]] in the sense that:
+::$(1): \quad f$ is an [[Definition:Injection|injection]]
+::$(2): \quad f$ is a [[Definition:Surjection|surjection]]
+{{iff}}:
+:the [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ of $f$ is itself a [[Definition:Mapping|mapping]].
+\end{theorem}
+
+\begin{proof}
+=== [[Inverse of Injective and Surjective Mapping is Mapping|Necessary Condition]] ===
+{{:Inverse of Injective and Surjective Mapping is Mapping}}
+=== [[Inverse is Mapping implies Mapping is Injection and Surjection|Sufficient Condition]] ===
+{{:Inverse is Mapping implies Mapping is Injection and Surjection}}
+\end{proof}
+
+\begin{proof}
+=== [[Inverse of Injective and Surjective Mapping is Mapping/Proof 2|Necessary Condition]] ===
+{{:Inverse of Injective and Surjective Mapping is Mapping/Proof 2}}{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f^{-1}: T \to S$ be a [[Definition:Mapping|mapping]].
+By [[Inverse Mapping is Bijection]], both $f$ and $f^{-1}$ are [[Definition:Bijection|bijections]].
+Hence, in particular, $f$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Composite Mapping to be Identity}
+Tags: Identity Mappings, Injections, Surjections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ and $g: T \to S$ be [[Definition:Mapping|mappings]] such that:
+: $g \circ f = I_S$
+where $I_S$ is the [[Definition:Identity Mapping|identity mapping]] on $S$.
+Then $f$ is an [[Definition:Injection|injection]] and $g$ is a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+From [[Identity Mapping is Bijection]], $I_S$ is a [[Definition:Bijection|bijection]].
+From [[Identity Mapping is Injection]], $I_S$ is an [[Definition:Injection|injection]].
+From [[Injection if Composite is Injection]] it follows that $f$ is an [[Definition:Injection|injection]].
+From [[Identity Mapping is Surjection]], $I_S$ is a [[Definition:Surjection|surjection]].
+From [[Surjection if Composite is Surjection]] it follows that $g$ is a [[Definition:Surjection|surjection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection iff exists Mapping which is Left and Right Inverse}
+Tags: Bijections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then $f$ is a [[Definition:Bijection|bijection]] {{iff}}:
+:there exists a [[Definition:Mapping|mapping]] $g: T \to S$ such that:
+::$g \circ f = I_S$
+::$f \circ g = I_T$
+:where $I_S$ and $I_T$ are the [[Definition:Identity Mapping|identity mappings]] on $S$ and $T$ respectively.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be a [[Definition:Bijection/Definition 4|bijection]].
+Then [[Definition:Universal Quantifier|for each]] $y \in T$ there exists [[Definition:Unique|one and only one]] $x \in S$ such that $\map f x = y$.
+That is, that there exists a [[Definition:Mapping|mapping]] $g: T \to S$ with the property that:
+: $\forall y \in T: \exists x \in S: \map g y = x$
+Let $y \in T$.
+Let $x = g \map g y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \map f {\map g y}
+      | c = 
+}}
+{{eqn | r = \map {f \circ g} y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \map g {\map f x}
+      | c = 
+}}
+{{eqn | r = \map {f \circ g} x
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $g$ has the property that:
+:$g \circ f = I_S$
+:$f \circ g = I_T$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose there exists a $g$ which satisfies the conditions on $f$.
+By [[Condition for Composite Mapping to be Identity]]:
+:$(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an [[Definition:Injection|injection]].
+:$(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a [[Definition:Surjection|surjection]].
+Thus, by definition, $f$ is a [[Definition:Bijection/Definition 1|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eigenvalues of G-Representation are Roots of Unity}
+Tags: 
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $\left({K, +, \cdot}\right)$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $V$ be a [[Definition:G-Module|$G$-module]] over $K$ (i.e. $V$ is a $K \left[{G}\right]$-module).
+Then $\forall g \in G$, the [[Definition:Eigenvalue|eigenvalues]] of the action by the vector $g \in K \left[{G}\right]$ on $V$ are [[Definition:Root of Unity|roots of unity]].
+\end{theorem}
+
+\begin{proof}
+Fix an arbitrary $g \in G$ and consider the corresponding vector $g \in K \left[{G}\right]$.
+Let $\lambda$ be an eigenvalue of $g$, that is $\lambda$ is an eigenvalue of the map in $\operatorname{Aut} \left({V}\right): \vec v \mapsto g \vec v$.
+Then by definition of an eigenvalue we have:
+:$\exists \vec v_\lambda \in V : g \vec v_\lambda = \lambda \vec v_\lambda$
+Let $n$ be the order of $g$ in $G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \vec v_\lambda
+      | r = g^n \vec v_\lambda
+      | c = as $g^n = e$ and $e$ acts trivially on $V$
+}}
+{{eqn | r = \lambda g^{n-1} \vec v_\lambda
+      | c = as $G$ acts [[Definition:Linear Group Action|linearly]] on $V$
+}}
+{{eqn | r = \vdots
+      | c = Continue exchanging $g^k \vec v_\lambda$ with $\lambda g^{k-1} \vec v_\lambda$
+}}
+{{eqn | r = \lambda^n \vec v_\lambda
+      | c = as $g^0 = e$ which acts trivially on $V$
+}}
+{{end-eqn}}
+{{explain|"as $g^n {{=}} e$ and $e$ acts trivially on $V$" - this needs to be linked to and / or proved}}
+{{explain|"as $G$ acts [[Definition:Linear Group Action|linearly]] on $V$" - this needs to be linked to and / or proved}}
+{{explain|"as $g^0 = e$ which acts trivially on $V$" - this needs to be linked to and / or proved}}
+Thus:
+:$\vec v_\lambda = \lambda^n \vec v_\lambda$
+which means:
+:$\lambda^n = 1$
+and thus by definition, $\lambda$ is an $n$th root of unity.
+{{qed}}
+la77w1qnzjsm0z4avytxiwx9t7fuetp
+\end{proof}<|endoftext|>
+\section{Character of Representations over C are Algebraic Integers}
+Tags: Module Theory, Group Theory, Complex Analysis
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $\chi$ be the [[Definition:Character (Representation Theory)|character]] of any [[Definition:G-Module|$\C \left[{G}\right]$-module]] $\left({V, \rho}\right)$.
+Then for all $g \in G$, it follows that $\chi \left({g}\right)$ is an [[Definition:Algebraic Integer|algebraic integer]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Character (Representation Theory)|character]]:
+:$\chi \left({g}\right) = \operatorname{Tr} \left({\rho_g}\right)$
+{{explain|$\operatorname{Tr} \left({\rho_g}\right)$, and indeed $\rho_g$ itself.}}
+where:
+: $\rho \in \hom \left({\C \left[{G}\right], \operatorname{Aut} \left({V}\right)}\right): \vec {e_g} \mapsto \rho_g$
+by definition.
+{{explain|The above definition is not stated in that form on {{ProofWiki}}. Link to $\hom \left({\C \left[{G}\right]$ and $\operatorname{Aut} \left({V}\right)}\right)$, clarify specifically what $\C \left[{G}\right]$ is, and define $\vec {e_g}$.}}
+Fix an arbitrary $g \in G$.
+Let $\left\vert{g}\right\vert$ denote the [[Definition:Order of Group Element|order]] of $g$.
+The [[Definition:Trace (Linear Algebra)|trace]] $\operatorname{Tr} \left({\rho_g}\right)$ of $\rho_g$ is defined as the sum of the [[Definition:Eigenvalue|eigenvalues]] of $\rho_g$.
+{{explain|The definition of Trace needs to be reviewed.}}
+From [[Eigenvalues of G-Representation are Roots of Unity]], we have that any [[Definition:Eigenvalue|eigenvalue]] $\lambda$ of $\rho_g$ is a [[Definition:Complex Roots of Unity|root of unity]] whose [[Definition:Order of Root of Unity|order]] is $\left\vert{g}\right\vert$.
+We have that $\lambda$ satisfies the [[Definition:Monic Polynomial|monic polynomial]] $x^{\left\vert{g}\right\vert} - 1$
+Hence we have that $\lambda$ is an [[Definition:Algebraic Integer|algebraic integer]].
+From [[Ring of Algebraic Integers]], we have that the sum of the [[Definition:Eigenvalue|eigenvalues]] is also an [[Definition:Algebraic Integer|algebraic integer]].
+Thus $\chi \left({g}\right)$ is an [[Definition:Algebraic Integer|algebraic integer]].
+{{qed}}
+[[Category:Module Theory]]
+[[Category:Group Theory]]
+[[Category:Complex Analysis]]
+ncn9awjawffn7x5umv4yc1k7pls6q23
+\end{proof}<|endoftext|>
+\section{Injection from Finite Set to Itself is Surjection}
+Tags: Injections, Surjections, Finite Sets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $f: S \to S$ be an [[Definition:Injection|injection]].
+Then $f$ is also a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in S$.
+We need to show that there exists $b \in S$ such that $a = \map f b$.
+Consider what happens when $f$ is applied repeatedly on $S$.
+Let $f^2$ denote $f \circ f$ and, generally, $f^n := f \circ f^{n-1}$.
+Consider the [[Definition:Sequence|sequence]] of [[Definition:Element|elements]] of $S$:
+:$a, \map f a, \map {f^2} a, \ldots$
+Because $S$ is a [[Definition:Finite Set|finite set]], there must be repetitions.
+That is, there must exist $r, s \in \N$ such that:
+:$\map {f^r} a = \map {f^s} a$
+where $r \ne s$.
+{{WLOG}}, assume $r > s$.
+$f$ is an [[Definition:Injection|injection]].
+Therefore by [[Composite of Injections is Injection]], $f^s$ is an [[Definition:Injection|injection]].
+By [[Injection iff Left Cancellable]], $f^s$ is [[Definition:Left Cancellable Mapping|left cancellable]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {f^r} a
+      | r = \map {f^s} a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f^s \circ f^{r - s} } a
+      | r = \map {f^s} a
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f^{r - s} } a
+      | r = a
+      | c = {{Defof|Left Cancellable Mapping}}
+}}
+{{end-eqn}}
+That is, $b = \map {f^{r - s - 1} } a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Codomain of Bijection is Domain of Inverse}
+Tags: Bijections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
+Let $f^{-1}: T \to S$ be the [[Definition:Inverse Mapping|inverse]] of $f$.
+Then the [[Definition:Domain of Mapping|domain]] of $f^{-1}$ equals the [[Definition:Codomain of Mapping|codomain]] of $f$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]]:
+:$\Dom f = S$ and $\Cdm f = T$
+:$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$
+That is:
+:$\Dom {f^{-1} } = T = \Cdm f$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Bijection is Codomain of Inverse}
+Tags: Bijections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
+Let $f^{-1}: T \to S$ be the [[Definition:Inverse Mapping|inverse]] of $f$.
+Then the [[Definition:Codomain of Mapping|codomain]] of $f^{-1}$ equals the [[Definition:Domain of Mapping|domain]] of $f$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]]:
+:$\Dom f = S$ and $\Cdm f = T$
+:$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$
+That is:
+:$\Cdm {f^{-1} } = S = \Dom f$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conditions for Uniqueness of Left Inverse Mapping}
+Tags: Left Inverse Mappings, Conditions for Uniqueness of Left Inverse Mapping
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]] such that $S \ne \O$.
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+Then a [[Definition:Left Inverse Mapping|left inverse mapping]] of $f$ is in general not [[Definition:Unique|unique]].
+[[Definition:Unique|Uniqueness]] occurs under either of two circumstances:
+:$(1): \quad S$ is a [[Definition:Singleton|singleton]]
+:$(2): \quad f$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+If $f$ is a [[Definition:Bijection|bijection]], then by definition $f$ is also a [[Definition:Surjection|surjection]].
+Then:
+:$T \setminus \Img f = \O$
+:and we have that $g = f^{-1}$.
+As $f^{-1}$ is uniquely defined $g$ is itself unique.
+{{qed|lemma}}
+If $S$ is a [[Definition:Singleton|singleton]] then there can only be one [[Definition:Mapping|mapping]] $g: T \to S$:
+:$\forall t \in T: \map g t = s$
+{{qed|lemma}}
+If $f$ is not a [[Definition:Bijection|bijection]], then as it is an [[Definition:Injection|injection]] it can not be a [[Definition:Surjection|surjection]].
+Then:
+:$T \setminus \Img f \ne \O$
+Let $t \in T \setminus \Img f$.
+We can now choose any $x_0 \in S$ such that $\map g t = x_0$.
+If $S$ is not a [[Definition:Singleton|singleton]], such an $x_0$ is not [[Definition:Unique|unique]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Surjection iff Right Inverse/Non-Uniqueness}
+Tags: Surjection iff Right Inverse
+
+\begin{theorem}
+A [[Definition:Right Inverse Mapping|right inverse]] of $f$ is in general not [[Definition:Unique|unique]].
+Uniqueness occurs {{iff}} $f$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+If $f$ is not an [[Definition:Injection|injection]] then:
+:$\exists y \in T: \exists x_1, x_2 \in S: \map f {x_1} = y = \map f {x_2}$
+Hence we have more than one choice in $\map {f^{-1} } {\set y}$ for how to map $\map g y$.
+That is, $\map g y$ is not [[Definition:Unique|unique]].
+This does not happen {{iff}} $f$ is an [[Definition:Injection|injection]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping reflects Preordering}
+Tags: Preorder Theory, Mapping Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let ${\precsim} \subseteq T \times T$ be a [[Definition:Preordering|preordering]] on $T$.
+Let $\RR$ be the [[Definition:Relation|relation]] defined on $S$ by the rule:
+:$x \mathrel \RR y \iff \map f x \precsim \map f y$
+Then $\RR$ is a [[Definition:Preordering|preordering]] on $S$.
+\end{theorem}
+
+\begin{proof}
+=== Reflexivity ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | o = \in
+      | r = T
+      | c = Definition of $f$
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | o = \precsim
+      | r = \map f x
+      | c = as $\precsim$ is a [[Definition:Preordering|preordering]] on $T$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \RR
+      | r = x
+      | c = Definition of $\RR$
+}}
+{{end-eqn}}
+Thus $\RR$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Transitivity ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \RR
+      | r = y
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = y
+      | o = \RR
+      | r = z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | o = \precsim
+      | r = \map f y
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \map f y
+      | o = \precsim
+      | r = \map f z
+      | c = Definition of $\RR$
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | o = \precsim
+      | r = \map f z
+      | c = as $\precsim$ is a [[Definition:Preordering|preordering]] on $T$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \RR
+      | r = z
+      | c = Definition of $\RR$
+}}
+{{end-eqn}}
+Thus $\RR$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So, by definition, $\RR$ is a [[Definition:Preordering|preordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Inequality}
+Tags: Real Analysis, Inequalities, Bernoulli's Inequality
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > -1$.
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+:$\paren {1 + x}^n \ge 1 + n x$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\paren {1 + x}^n \ge 1 + nx$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+:$\paren {1 + x}^0 \ge 1$
+so $\map P 0$ holds.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\left({1 + x}\right)^k \ge 1 + kx$
+We need to show that:
+:$\paren {1 + x}^{k + 1} \ge 1 + \paren {k + 1} x$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + x}^{k + 1}
+      | r = \paren {1 + x}^k \paren {1 + x}
+}}
+{{eqn | o = \ge
+      | r = \paren {1 + k x} \paren {1 + x}
+      | c = [[Bernoulli's Inequality/Proof 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 1 + \paren {k + 1} x + k x^2
+}}
+{{eqn | o = \ge
+      | r = 1 + \paren {k + 1} x
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $y = 1 + x$.
+Then $y \ge 0$, and:
+:$\paren {1 + x}^n = 1 + \paren {y^n - 1}$
+If $y \ge 1$, then by [[Sum of Geometric Sequence]]:
+:$\displaystyle y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$
+If $y < 1$, then by [[Sum of Geometric Sequence]]:
+:$\displaystyle y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0}^{n - 1} y^k \ge -n \paren {1 - y} = n x$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Categories of Elements of Ring}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+The [[Definition:Element|elements]] of $R$ are [[Definition:Partition (Set Theory)|partitioned]] into three classes:
+:$(1): \quad$ the [[Definition:Ring Zero|zero]]
+:$(2): \quad$ the [[Definition:Unit of Ring|units]]
+:$(3): \quad$ the [[Definition:Proper Element of Ring|proper elements]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Proper Element of Ring|proper element]] is a non-[[Definition:Ring Zero|zero]] element which has no [[Definition:Ring Product Inverse|product inverse]].
+Also by definition, a [[Definition:Unit of Ring|unit]] is an element which does have a [[Definition:Ring Product Inverse|product inverse]].
+Because $0 \circ x = 0$ there can be no $x \in R$ such that $0 \times x = 1$, and so $0$ is not a [[Definition:Unit of Ring|unit]].
+Hence the result.
+{{qed}}
+[[Category:Ring Theory]]
+76cxzcd0t4nheq7khu1x1x3rp3cc4sw
+\end{proof}<|endoftext|>
+\section{Multiple of Ring Product}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $x, y \in \left({R, +, \circ}\right)$.
+Then:
+:$\forall n \in \Z_{> 0}: \left({n \cdot x} \right) \circ y = n \cdot \left({x \circ y}\right) = x \circ \left({n \cdot y}\right)$
+where $n \cdot x$ denotes the [[Definition:Integral Multiple/Rings and Fields|$n$th multiple]] of $x$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\displaystyle n \cdot x := \sum_{j \mathop = 1}^n x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \left({n \cdot x} \right) \circ y
+      | r = \left({\sum_{j \mathop = 1}^n x} \right) \circ y
+      | c = by definition of [[Definition:Integral Multiple/Rings and Fields|integral multiple]]
+}}
+{{eqn | n = 1
+      | r = \sum_{j \mathop = 1}^n \left({x \circ y} \right)
+      | c = [[General Distributivity Theorem]]
+}}
+{{eqn | r = n \cdot \left({x \circ y}\right)
+      | c = by definition of [[Definition:Integral Multiple/Rings and Fields|integral multiple]]
+}}
+{{eqn | r = x \circ \left({\sum_{j \mathop = 1}^n y} \right)
+      | c = [[General Distributivity Theorem]] from $(1)$
+}}
+{{eqn | r = x \circ \left({n \cdot y}\right)
+      | c = by definition of [[Definition:Integral Multiple/Rings and Fields|integral multiple]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ring Theory]]
+g02rj9281x192zcs329rxqfssoa0b8q
+\end{proof}<|endoftext|>
+\section{Ordering is Preserved on Integers by Addition}
+Tags: Integer Addition, Orderings on Integers
+
+\begin{theorem}
+The [[Definition:Usual Ordering|usual ordering]] on the [[Definition:Integer|integers]] is preserved by the operation of [[Definition:Integer Addition|addition]]:
+:$\forall a, b, c, d, \in \Z: a \le b, c \le d \implies a + c \le b + d$
+\end{theorem}
+
+\begin{proof}
+Recall that [[Integers form Ordered Integral Domain]].
+Then from [[Relation Induced by Strict Positivity Property is Compatible with Addition]]:
+:$\forall x, y, z \in \Z: x \le y \implies \paren {x + z} \le \paren {y + z}$
+:$\forall x, y, z \in \Z: x \le y \implies \paren {z + x} \le \paren {z + y}$
+So:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \le
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + c
+      | o = \le
+      | r = b + c
+      | c = [[Relation Induced by Strict Positivity Property is Compatible with Addition]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \le
+      | r = d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b + c
+      | o = \le
+      | r = b + d
+      | c = [[Relation Induced by Strict Positivity Property is Compatible with Addition]]
+}}
+{{end-eqn}}
+Finally:
+{{begin-eqn}}
+{{eqn | l = a + c
+      | o = \le
+      | r = b + c
+}}
+{{eqn | l = b + c
+      | o = \le
+      | r = b + d
+}}
+{{eqn | ll= \leadsto
+      | l = a + c
+      | o = \le
+      | r = b + d
+      | c = {{Defof|Ordering}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set of Finite Set is Finite}
+Tags: Power Set
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Then the [[Definition:Power Set|power set]] of $S$ is likewise [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Then by definition:
+:$\exists n \in \N: \card S = n$
+where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
+From [[Cardinality of Power Set of Finite Set]]:
+:$\card {\powerset S} = 2^n$
+where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+As $n \in \N$ it follows that $2^n \in \N$ and so $\powerset S$ is also by definition a [[Definition:Finite Set|finite set]].
+{{qed}}
+[[Category:Power Set]]
+tikj18ioh81t6cn77m6ng7119sxrh4b
+\end{proof}<|endoftext|>
+\section{Common Divisor Divides Difference}
+Tags: Divisors, Common Divisor Divides Difference
+
+\begin{theorem}
+Let $c$ be a [[Definition:Common Divisor of Integers|common divisor]] of two [[Definition:Integer|integers]] $a$ and $b$.
+That is:
+:$a, b, c \in \Z: c \divides a \land c \divides b$
+Then:
+:$c \divides \paren {a - b}$
+\end{theorem}
+
+\begin{proof}
+Let $c \divides a \land c \divides b$.
+From [[Common Divisor Divides Integer Combination]]:
+:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
+Putting $p = 1$ and $q = -1$:
+:$c \divides \paren {a - b}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \divides
+      | r = a
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x \in \Z:
+      | l = a
+      | r = x c
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | l = c
+      | o = \divides
+      | r = b
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists y \in \Z:
+      | l = b
+      | r = y c
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | ll= \leadsto
+      | l = a - b
+      | r = x c - y c
+      | c = substituting for $a$ and $b$
+}}
+{{eqn | r = \paren {x - y} c
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists z \in \Z:
+      | l = a - b
+      | r = z c
+      | c = where $z = x - y$
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | o = \divides
+      | r = \paren {a - b}
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisor Divides Multiple}
+Tags: Divisors, Divisor Divides Multiple
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Integer|integers]].
+Let:
+:$a \divides b$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\forall c \in \Z: a \divides b c$
+\end{theorem}
+
+\begin{proof}
+We have that [[Integers form Integral Domain]].
+The result then follows from [[Multiple of Divisor in Integral Domain Divides Multiple]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.
+Then:
+:$\paren {a d} c = b c$
+that is:
+:$\paren {a c} d = b c$
+which follows because [[Integer Multiplication is Commutative]] and [[Integer Multiplication is Associative]].
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a \divides b$.
+From [[Integer Divides Zero]]:
+:$a \divides 0$
+Thus $a$ is a [[Definition:Common Divisor of Integers|common divisor]] of $b$ and $0$.
+From [[Common Divisor Divides Integer Combination]]:
+:$\forall p, q \in \Z: a \divides \paren {p \cdot b + q \cdot 0}$
+Putting $p = c$ and $q = 1$ (for example):
+:$a \divides \paren {c b + 0}$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \divides
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x \in \Z:
+      | l = b
+      | r = x a
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | ll= \leadsto
+      | l = b c
+      | r = x c a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists z \in \Z:
+      | l = b c
+      | r = z a
+      | c = where $z = x c$
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = \divides
+      | r = b c
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Integer has Unique Positive Integer Associate}
+Tags: Integers
+
+\begin{theorem}
+Let $a \in \Z$ be an [[Definition:Integer|integer]] such that $a \ne 0$.
+Then $a$ has a [[Definition:Unique|unique]] [[Definition:Associate of Integer|associate]] $b \in \Z_{>0}$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in \Z_{\ne 0}$.
+By [[Integer Divides its Absolute Value]]:
+:$a \divides \size a$ and $\size a \divides a$
+Hence $\size a$ is an [[Definition:Associate of Integer|associate]] of $a$.
+Now we prove its [[Definition:Unique|uniqueness]].
+Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$.
+Let $a \sim b$ and $a \sim c$ where $\sim$ denotes the relation of [[Definition:Associate of Integer|associatehood]].
+By definition of [[Definition:Associate of Integer|associatehood]]:
+:$a \divides b$ and $b \divides a$
+and:
+:$a \divides c$ and $c \divides a$
+From [[Divisor Relation is Antisymmetric/Corollary/Proof 2]]:
+:$a = \pm b$
+and
+:$a = \pm c$
+That is:
+:$\pm b = \pm c$
+which means:
+:$b = c$ or $b = -c$
+But as both $b > 0$ and $c > 0$:
+:$b = c$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One is not Prime}
+Tags: Prime Numbers, 1, One is not Prime
+
+\begin{theorem}
+The [[Definition:Integer|integer]] [[1|$1$ (one)]] is not a [[Definition:Prime Number|prime number]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Prime Number/Definition 1|prime number]] is a [[Definition:Positive Integer|positive integer]] which has exactly $2$ [[Definition:Divisor of Integer|divisors]] which are themselves [[Definition:Positive Integer|positive integers]].
+From [[Divisors of One]], the only [[Definition:Divisor of Integer|divisors]] of $1$ are $1$ and $-1$.
+So the only [[Definition:Divisor of Integer|divisor]] of $1$ which is a [[Definition:Positive Integer|positive integer]] is $1$.
+As $1$ has only one such [[Definition:Divisor of Integer|divisor]], it is not classified as a [[Definition:Prime Number|prime number]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sigma Function of Prime Number]], the [[Definition:Integer Addition|sum]] $\map \sigma p$ of all the [[Definition:Positive Integer|positive integer]] [[Definition:Divisor of Integer|divisors]] of a [[Definition:Prime Number|prime number]] $p$ is $p + 1$.
+But from [[Sigma Function of 1]], $\map \sigma 1 = 1$.
+If $1$ were to be classified as [[Definition:Prime Number|prime]], then $\map \sigma 1$ would be an exception to the rule that $\map \sigma p = p + 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Initial Part of String is Substring}
+Tags: Formal Systems
+
+\begin{theorem}
+Let $S$ be a [[Definition:String|string]], and let $T$ be an [[Definition:Initial Part|initial part]] of $S$.
+Then $T$ is a [[Definition:Substring|substring]] of $S$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Substring|substring]], there exists a [[Definition:String|string]] $T'$ such that:
+:$S = TT'$
+Hence $S$ is the [[Definition:Concatenation (Formal Systems)|concatenation]] of the [[Definition:Null String|null string]], $T$, and $T'$.
+Thus by definition of [[Definition:Substring|substring]], $T$ is a [[Definition:Substring|substring]] of $S$.
+{{qed}}
+[[Category:Formal Systems]]
+brdmwx4fr7kw0ci47xvs1zgz98lmxep
+\end{proof}<|endoftext|>
+\section{Ring Zero is Idempotent}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|ring zero]] is $0_R$.
+Then $0_R$ is an [[Definition:Idempotent Element|idempotent element]] of $R$ under the [[Definition:Ring Product|ring product]] $\circ$:
+:$0_R \circ 0_R = 0_R$
+\end{theorem}
+
+\begin{proof}
+By [[Ring Product with Zero]] (applied to $0_R$):
+:$0_R \circ 0_R = 0_R$
+which was to be proven. 
+{{qed}}
+[[Category:Ring Theory]]
+ss3g8gf6liesnsozzglyn9w1zk3odym
+\end{proof}<|endoftext|>
+\section{Prime iff Coprime to all Smaller Positive Integers}
+Tags: Prime Numbers, Coprime Integers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]]
+Then:
+:$\forall x \in \Z, 0 < x < p: x \perp p$
+That is, $p$ is [[Definition:Coprime Integers|relatively prime]] to all smaller [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+\end{theorem}
+
+\begin{proof}
+From [[Prime not Divisor implies Coprime]], if $p$ does not [[Definition:Divisor of Integer|divide]] an [[Definition:Integer|integer]] $x$, it is [[Definition:Coprime Integers|relatively prime]] to $x$.
+From [[Absolute Value of Integer is not less than Divisors/Corollary|Absolute Value of Integer is not less than Divisors: Corollary]], $p$ does not [[Definition:Divisor of Integer|divide]] an [[Definition:Integer|integer]] smaller than $p$.
+It follows that $p$ is [[Definition:Coprime Integers|relatively prime]] to all smaller [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+The special case when $x = 0$ is excluded as from [[Integers Coprime to Zero]], $p$ is not [[Definition:Coprime Integers|relatively prime]] to $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Positive Dividend}
+Tags: Division Theorem
+
+\begin{theorem}
+<onlyinclude>
+For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Unique|unique]] integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+<onlyinclude>
+In the above equation:
+* $a$ is the '''dividend'''
+* $b$ is the '''divisor'''
+* $q$ is the '''quotient'''
+* $r$ is the '''principal remainder''', or, more usually, just the '''remainder'''.
+\end{theorem}
+
+\begin{proof}
+This result can be split into two parts:
+=== [[Division Theorem/Positive Divisor/Positive Dividend/Existence|Proof of Existence]] ===
+{{:Division Theorem/Positive Divisor/Positive Dividend/Existence}}
+=== [[Division Theorem/Positive Divisor/Positive Dividend/Uniqueness|Proof of Uniqueness]] ===
+{{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness}}
+[[Category:Division Theorem]]
+5mooiyv6ltjxhds9a5ijyjvoog14knh
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor}
+Tags: Division Theorem
+
+\begin{theorem}
+<onlyinclude>
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+<onlyinclude>
+In the above equation:
+:$a$ is the '''dividend'''
+:$b$ is the '''divisor'''
+:$q$ is the '''quotient'''
+:$r$ is the '''principal remainder''', or, more usually, just the '''remainder'''.
+\end{theorem}
+
+\begin{proof}
+This result can be split into two parts:
+=== [[Division Theorem/Positive Divisor/Existence|Proof of Existence]] ===
+{{:Division Theorem/Positive Divisor/Existence}}
+=== [[Division Theorem/Positive Divisor/Uniqueness|Proof of Uniqueness]] ===
+{{:Division Theorem/Positive Divisor/Uniqueness}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Positive Dividend/Existence}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Positive Dividend/Uniqueness}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $a \ge 0$ and $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
+:$\forall a, b \in \Z, a \ge 0, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Existence/Proof 1}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+From [[Division Theorem/Positive Divisor/Positive Dividend/Existence|Division Theorem: Positive Divisor: Positive Dividend: Existence]]:
+:$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+That is, the result holds for [[Definition:Positive Integer|positive]] $a$.
+{{qed|lemma}}
+It remains to be shown that the statement holds for $a < 0$.
+From [[Division Theorem/Positive Divisor/Positive Dividend|Division Theorem: Positive Divisor: Positive Dividend]]:
+:$\exists \tilde q, \tilde r \in \Z: \size a = \tilde q b + \tilde r, 0 \le \tilde r < b$
+where $\size a$ denotes the [[Definition:Absolute Value|absolute value]] of $a$: by definition $\size a \ge 0$.
+As $a < 0$ it follows by definition of [[Definition:Absolute Value|absolute value]] that $\size a = -a$.
+This gives:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = -\size a
+      | c = 
+}}
+{{eqn | r = -\paren {\tilde q b + \tilde r}
+      | c = 
+}}
+{{eqn | r = b \paren {-\tilde q} + \paren {-\tilde r}
+      | c = 
+}}
+{{end-eqn}}
+Let $\tilde r = 0$.
+Then:
+: $q = -\tilde q, r = \tilde r = 0$
+which gives:
+: $a = q b + r, 0 \le r < b$
+as required.
+Otherwise:
+:$0 < \tilde r < b \implies 0 < b - \tilde r < b$
+which suggests a rearrangement of the expression for $a$ above:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = b \paren {-\tilde q} + \paren {-\tilde r}
+      | c = 
+}}
+{{eqn | r = b \paren {-\tilde q} - b + b - \tilde r
+      | c = 
+}}
+{{eqn | r = b \paren {-1 - \tilde q} + \paren {b - \tilde r}
+      | c = 
+}}
+{{end-eqn}}
+Taking:
+: $q = \paren {-1 - \tilde q}$
+and:
+: $r = \paren {b - \tilde r}$
+the required result follows.
+[[Definition:Unique|Uniqueness]] of $q$ and $r$ follow from the [[Definition:Unique|Uniqueness]] of $\tilde q$ and $\tilde r$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Existence/Proof 2}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+Let:
+: $q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$
+where $\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]].
+Thus $q \in \Z$ and $t \in \hointr 0 1$.
+So:
+:$\dfrac a b = q + t$
+and so:
+:$(1): \quad a = q b + r$
+where $r = t d$.
+Since $a, q, b \in \Z$, it follows from $(1)$ that:
+:$r = a - q b$
+and so $r \in \Z$ also.
+Since $0 \le t < 1$ and $b > 0$, it follows that:
+:$0 \le t b < b$
+that is:
+:$0 \le r < b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Existence}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Uniqueness/Proof 1}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
+{{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Uniqueness/Proof 2}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
+Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$.
+Thus:
+:$\dfrac a b = q + \dfrac r b$
+and:
+:$0 \le \dfrac r b \le \dfrac {b - 1} b < 1$
+So:
+:$q = \floor {\dfrac a b}$
+and so:
+:$r = a - b \floor {\dfrac a b}$
+Thus, given $a$ and $b$, the numbers $q$ and $r$ are [[Definition:Unique|unique]] determined.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Uniqueness}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Existence/Proof 3}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, there exist [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$:
+:$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+Let there exist $q \in Z$ such that $a - b q = 0$.
+Then $a = b q$ as required, with $r = 0$.
+Otherwise, let $S$ be defined as the [[Definition:Set|set]] of all [[Definition:Positive Integer|positive integers]] of the form $a - z b$ where $z$ is an [[Definition:Integer|integer]]:
+:$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$
+Setting $z = 0$ it is seen that $a \in S$, so $S \ne \O$.
+From [[Set of Integers Bounded Below by Integer has Smallest Element]], $S$ has a [[Definition:Smallest Element|smallest element]].
+Let $r$ be the [[Definition:Smallest Element|smallest element]] of $S$.
+Let $r = a - b q$.
+As there does not exist $q \in Z$ such that $a - b q = 0$:
+:$r > 0$
+Suppose $r = b$.
+Then $a = b \paren {q + 1}$ and it has already been declared that there does not exist such a $q + 1 \in Z$.
+Suppose $r > b$.
+Then $x = a - b \paren {q + 1} \in S$ such that $x < r$, which contradicts the assumption that $r$ is the [[Definition:Smallest Element|smallest element]] of $S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem/Positive Divisor/Uniqueness/Proof 3}
+Tags: Division Theorem
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b > 0$, the [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < b$ are [[Definition:Unique|unique]]:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+\end{theorem}
+
+\begin{proof}
+It is given by [[Division Theorem/Positive Divisor/Existence|Division Theorem: Positive Divisor: Existence]] that such $q$ and $r$ exist.
+Suppose that:
+:$a = b q_1 + r_1 = b q_2 + r_2$
+where both $0 \le r_1 < b$ and $0 \le r_2 < b$.
+{{WLOG}}, suppose $r_1 \ge r_2$.
+Then:
+:$r_1 - r_2 = b \paren {q_2 - q_1}$
+That is:
+:$b \divides \paren {r_2 - r_1}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+But:
+:$r_1 - r_2 < b$
+while from [[Absolute Value of Integer is not less than Divisors/Corollary|Absolute Value of Integer is not less than Divisors: Corollary]]:
+:$r_1 - r_2 \ge b$
+unless from [[Integer Divides Zero]] $r_1 - r_2 = 0$.
+So $r_1 = r_2$ and it follows directly that $q_1 = q_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Coprime to all Factors is Coprime to Whole}
+Tags: Coprime Integers, Integer Coprime to all Factors is Coprime to Whole
+
+\begin{theorem}
+Let $a, b \in \Z$.
+Let $\displaystyle b = \prod_{j \mathop = 1}^r b_j$
+Let $a$ be [[Definition:Coprime Integers|coprime]] to each of $b_1, \ldots, b_r$.
+Then $a$ is [[Definition:Coprime Integers|coprime]] to $b$.
+\end{theorem}
+
+\begin{proof}
+From [[Integer Combination of Coprime Integers]]:
+:$\forall j \in \set {1, 2, \ldots, r}: a x_j + b_j y_j = 1$
+for some $x_j, y_j \in \Z$.
+Thus:
+:$\displaystyle \prod_{j \mathop = 1}^r b_j y_j = \prod_{j \mathop = 1}^r \paren {1 - a x_j}$
+But $\displaystyle \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ is of the form $1 - a z$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \prod_{j \mathop = 1}^r b_j \prod_{j \mathop = 1}^r y_j
+      | r = 1 - a z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a z + b y
+      | r = 1
+      | c = where $\displaystyle y = \prod_{j \mathop = 1}^r y_j$
+}}
+{{end-eqn}}
+and so, by [[Integer Combination of Coprime Integers]], $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+''This proof follows the structure of {{AuthorRef|Euclid}}'s proof, if not its detail.''
+Let $a, b, c \in \Z$ such that $c$ is [[Definition:Coprime Integers|coprime]] to each of $a$ and $b$.
+Let $d = a b$.
+{{AimForCont}} $c$ and $d$ are not [[Definition:Coprime Integers|coprime]].
+Then:
+:$\exists e \in \Z_{>1}: e \divides c, e \divides d$
+We have that $c \perp a$ and $e \divides c$
+From {{EuclidPropLink|book=VII|prop=23|title=Divisor of One of Coprime Numbers is Coprime to Other}}:
+:$e \perp a$
+As $e \divides d$ it follows by definition of [[Definition:Divisor of Integer|divisibility]] that:
+:$\exists f \in \Z: e f = d$
+We also have that:
+:$a b = d$
+So:
+:$a b = e f$
+But from {{EuclidPropLink|book=VII|prop=19|title=Relation of Ratios to Products}}:
+:$e : a = b : f$
+or in more contemporary language:
+:$\dfrac a e = \dfrac b f$
+But $a \perp e$.
+From {{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}, $\dfrac a e$ is in [[Definition:Canonical Form of Rational Number|canonical form]].
+From {{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}}:
+:$e \divides b$
+But we already have that:
+:$e \divides c$
+So $e$ is a [[Definition:Divisor of Integer|divisor]] of both $b$ and $c$.
+But this is a [[Proof by Contradiction|contradiction]] of the assertion that $c$ and $b$ are [[Definition:Coprime Integers|coprime]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{All Factors Divide Integer then Whole Divides Integer}
+Tags: Coprime Integers
+
+\begin{theorem}
+Let $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of the [[Definition:Integer|integers]].
+Let $S$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]].
+Let:
+:$\forall j \in \set {1, 2, \ldots, r}: a_r \divides b$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\displaystyle \prod_{j \mathop = 1}^r a_j \divides b$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is [[Definition:Pairwise Coprime Integers|pairwise coprime]].
+For all $r \in \N_{> 1}$, let $\map P r$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \prod_{j \mathop = 1}^r a_j \divides b$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+:$a_1 a_2 \divides b$
+By definition of [[Definition:Divisor of Integer|divisibility]]:
+:$b = a_1 q_1 = a_2 q_2$
+where $q_1, q_2 \in \Z$.
+By [[Integer Combination of Coprime Integers]]:
+:$\exists x, y \in \Z: a_1 x + a_2 y = 1$
+So:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = a_2 q_2 \cdot a_1 x + a_1 q_1 \cdot a_2 y
+      | c = 
+}}
+{{eqn | r = a_1 a_2 \paren {q_2 x + q_1 y}
+      | c = 
+}}
+{{end-eqn}}
+and so $a_1 a_2 \divides b$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$
+Then we need to show:
+:$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+[[Definition:By Hypothesis|By hypothesis]]:
+:$a_{k + 1} \divides b$
+By [[Integer Coprime to all Factors is Coprime to Whole]]:
+:$\displaystyle a_{k + 1} \perp \prod_{j \mathop = 1}^k a_j$
+By the [[All Factors Divide Integer then Whole Divides Integer#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$
+So by the [[All Factors Divide Integer then Whole Divides Integer#Basis for the Induction|basis for the induction]]:
+:$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+$\displaystyle r \in \N_{> 1}: \prod_{j \mathop = 1}^r a_j \divides b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Piecewise Continuous Function with One-Sided Limits is Bounded}
+Tags: Piecewise Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $f$ be a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]].
+Then $f$ is a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]].
+By definition, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
+For every $i \in \set {1, 2, \ldots, n}$, we define a [[Definition:Real Function|function]] $f_i$ with [[Definition:Domain of Mapping|domain]] $\closedint {x_{i − 1} } {x_i}$, as follows:
+:$\map {f_i} x := \begin{cases}
+\displaystyle \lim_{x \mathop \to x_{i − 1}^+} \map f x & : x = x_{i - 1} \\
+\map f x & : x \in \openint {x_{i - 1} } {x_i} \\
+\displaystyle \lim_{x \mathop \to x_i^-} \map f x & : x = x_i
+\end{cases}$
+The [[Definition:One-Sided Limit of Real Function|one-sided limits]] in this definition exist because $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]].
+We have that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$.
+We also have that $f_i = f$ on $\openint {x_{i - 1} } {x_i}$.
+Therefore $f_i$ is also [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$.
+By definition, $f_i$ is [[Definition:Right-Continuous at Point|right-continuous]] at $x_{i − 1}$ and [[Definition:Left-Continuous at Point|left-continuous]] at $x_i$.
+Therefore, $f_i$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] throughout its [[Definition:Domain of Mapping|domain]] $\closedint {x_{i − 1} } {x_i}$.
+By [[Continuous Function on Compact Subspace of Euclidean Space is Bounded]] and [[Closed Real Interval is Compact in Metric Space]], $f_i$ is [[Definition:Bounded Real-Valued Function|bounded]].
+Therefore, $f_i$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{i − 1} } {x_i}$.
+We have that $\openint {x_{i - 1} } {x_i}$ constitutes a [[Definition:Subset|subset]] of $\closedint {x_{i − 1} } {x_i}$.
+Thus $f_i$ is also bounded on $\openint {x_{i - 1} } {x_i}$.
+As $f_i = f$ on $\openint {x_{i - 1} } {x_i}$, $f$ is also [[Definition:Bounded Real-Valued Function|bounded]] on $\openint {x_{i - 1} } {x_i}$.
+We have that:
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the [[Definition:Open Real Interval|intervals]] $\openint {x_{i - 1} } {x_i}$
+and
+:the [[Definition:Set|set]] of these [[Definition:Open Real Interval|intervals]] is [[Definition:Finite Set|finite]].
+Hence the [[Definition:Set|set]] of [[Definition:Bounded Real-Valued Function|bounds]] of $f$ on these [[Definition:Open Real Interval|intervals]] is itself [[Definition:Bounded Subset of Real Numbers|bounded]].
+The [[Definition:Bound of Subset of Real Numbers|bound]] of this [[Definition:Set|set]] of [[Definition:Bound of Subset of Real Numbers|bounds]] serves as a [[Definition:Bound of Real-Valued Function|bound]] for $f$ on each of the [[Definition:Open Real Interval|intervals]] $\openint {x_{i - 1} } {x_i}$.
+Therefore, this [[Definition:Bound of Subset of Real Numbers|bound]] is a [[Definition:Bound of Real-Valued Function|bound]] for $f$ on the [[Definition:Set Union|union]] of these [[Definition:Open Real Interval|intervals]].
+That is, $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the [[Definition:Set Union|union]] of $\openint {x_{i - 1} } {x_i}$ for all $i \in \set {1, 2, \ldots, n}$.
+The only points left to consider are the points in the set $\set {x_0, x_1, \ldots, x_n}$. 
+Since this set is [[Definition:Finite Set|finite]], the [[Definition:Max Function|maximum]] $\map \max {\size {\map f {x_0} }, \size {\map f {x_1} }, \ldots, \size {\map f {x_n} } }$ is finite and serves as a [[Definition:Bounded Real-Valued Function|bound]] for $f$ on $\set {x_0, x_1, \ldots, x_n}$.
+Thus $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the whole of its [[Definition:Domain of Mapping|domain]] $\closedint a b$.
+Hence $f$ is a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bounds for Prime Numbers/Result 1}
+Tags: Upper Bounds for Prime Numbers
+
+\begin{theorem}
+:$\forall n \in \N: \map p n \le 2^{2^{n - 1} }$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Second Principle of Mathematical Induction|strong induction]]:
+Let us write $p_n = \map p n$.
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\map p n \le 2^{2^{n - 1} }$
+=== Basis for the Induction ===
+$\map P 1$ is true, as this just says $2 \le 2^{2^0} = 2$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Suppose that, for some $k \in \N$, each of $\map P 1, \map P 2, \ldots, \map P k$ is true.
+It remains to show that it logically follows that $\map P {k + 1}$ is true.
+That is, that:
+:$\map p {k + 1} \le 2^{2^k}$
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \map p {k + 1}
+      | o = \le
+      | r = p_1 p_2 \cdots p_k + 1
+      | c = [[Euclid's Theorem]]
+}}
+{{eqn | o = \le
+      | r = 2 \times 2^2 \times 2^{2^2} \times \cdots \times 2^{2^{k - 1} } + 1
+      | c = [[Upper Bounds for Prime Numbers/Result 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 2^{1 + 2 + 2^2 + 2^3 + \cdots + 2^{k - 1} } + 1
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = 2^{2^k - 1} + 1
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | o = \le
+      | r = 2^{2^k - 1} + 2^{2^k - 1}
+      | c = since $1 \le 2^{2^k - 1}$ for all $k$
+}}
+{{eqn | r = 2^{2^k}
+      | c = 
+}}
+{{end-eqn}}
+The result follows by the [[Second Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bounds for Prime Numbers/Result 2}
+Tags: Upper Bounds for Prime Numbers
+
+\begin{theorem}
+:$\forall n \in \N: \map p n \le \paren {p \paren {n - 1} }^{n - 1} + 1$
+\end{theorem}
+
+\begin{proof}
+Let us write $p_n = \map p n$.
+Let us take $N = p_1 p_2 \cdots p_n + 1$.
+By the same argument as in [[Euclid's Theorem]], we have that either $N$ is [[Definition:Prime Number|prime]], or it is not.
+If $N$ is [[Definition:Prime Number|prime]], then either $N = p_{n + 1}$ or not, in which case $N > p_{n + 1}$.
+In the second case, $N$ has a [[Definition:Prime Factor|prime factor]] not in $\set {p_1, p_2, \ldots, p_n}$
+Therefore it must have a [[Definition:Prime Factor|prime factor]] greater than any of $\set {p_1, p_2, \ldots, p_n}$.
+In any case, the next [[Definition:Prime Number|prime]] after $p_n$ can be no greater than $p_1 p_2 \cdots p_n + 1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bounds for Prime Numbers/Result 3}
+Tags: Upper Bounds for Prime Numbers
+
+\begin{theorem}
+: $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$
+\end{theorem}
+
+\begin{proof}
+Let us write $p_n = p \left({n}\right)$.
+From [[Bertrand's Conjecture]], for each $n \ge 2$ there exists a [[Definition:Prime Number|prime]] $p$ such that $n < p < 2 n$.
+For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+: $p_n < 2^n$
+$P(1)$ is the statement:
+:$p_1 = 2 = 2^1$
+As this does not fulfil the criterion:
+:$p \left({n}\right) < 2^n$
+it is not included in the result.
+=== Basis for the Induction ===
+$P(2)$ is true, as this just says:
+:$p_2 = 3 < 2^2 = 4$
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$p_k < 2^k$
+Then we need to show:
+:$p_{k+1} < 2^{k+1}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Suppose that $p_k < 2^k$ for some $k \ge 2$.
+By [[Bertrand's Conjecture]], there exists a prime $q$ such that $2^k < q < 2^{k+1}$.
+So:
+: $p < 2^k < q < 2^{k+1}$
+But as $q$ is a prime exceeding $p_k$, it means:
+: $p_{k+1} \le q < 2^{k+1}$
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+: $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$
+{{qed}}
+[[Category:Upper Bounds for Prime Numbers]]
+gshqsisc4nz55ru10a6lfuqs81lt1vg
+\end{proof}<|endoftext|>
+\section{Product of Integers of form 4n + 1}
+Tags: Arithmetic, Number Theory
+
+\begin{theorem}
+Let $m, n \in \Z$ such that both $m$ and $n$ are of the form $4 k + 1$ where $k \in \Z$.
+Then $m n$ is also of the form $4 k + 1$.
+\end{theorem}
+
+\begin{proof}
+Let $m = 4 k_1 + 1, n = 4 k_2 + 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = m n
+      | r = \paren {4 k_1 + 1} \paren {4 k_2 + 1}
+      | c = 
+}}
+{{eqn | r = 4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2 + 1
+      | c = multiplying out
+}}
+{{eqn | r = 4 \paren {4 k_1 k_2 + k_1 + k_2} + 1
+      | c = factoring $4$
+}}
+{{eqn | r = 4 k + 1
+      | c = where $k = 4 k_1 k_2 + k_1 + k_2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Number of Primes of form 4n - 1}
+Tags: Arithmetic, Number Theory, Prime Numbers
+
+\begin{theorem}
+There are [[Definition:Infinite Set|infinitely many]] [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} the contrary.
+That is, suppose there is a [[Definition:Finite Set|finite number]] of [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$.
+Let there be $k$ of them: $p_1, p_2, \ldots, p_k$.
+Let $S = \set {p_1, p_2, \ldots, p_k}$.
+Let $N$ be constructed as:
+:$\displaystyle N = 4 \prod_{i \mathop = 1}^k p_i - 1$
+If $N$ is a [[Definition:Prime Number|prime number]], then it is of the form $4 n - 1$.
+But then we have that $N \notin S$, which means that $S$ is not the complete set of [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$.
+Therefore $N$ must be [[Definition:Composite Number|composite]].
+Suppose all the [[Definition:Prime Factor|prime factors]] of $N$ are of the form $4 n + 1$.
+Then from [[Product of Integers of form 4n + 1|Product of Integers of form $4 n + 1$]] it follows that $N$ itself is of the form $4 n + 1$.
+Therefore $N$ must have at least one [[Definition:Prime Factor|prime factor]] $p$ of the form $4 n - 1$.
+Suppose $p \in S$.
+We have that:
+: $\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i$
+and so:
+: $\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i - p$
+So:
+:$N = q p + \left({p - 1}\right)$
+where:
+:$q = \dfrac {\displaystyle 4 \prod_{i \mathop = 1}^k p_i - p} p$
+So applying the [[Euclidean Algorithm]]:
+{{begin-eqn}}
+{{eqn | l = \gcd \set {N, p}
+      | r = \gcd \set {p - 1, p}
+      | c = 
+}}
+{{eqn | r = \gcd \set {p - 1, 1}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+So if $p \in S$, then it cannot be a [[Definition:Divisor of Integer|divisor]] of $N$.
+Therefore there must be a [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$ which is not in $S$.
+Therefore $S$ is not the complete set of [[Definition:Prime Number|prime numbers]] of the form $4 n - 1$.
+Therefore the assumption that $S$ is [[Definition:Finite Set|finite]] was [[Definition:False|false]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Coprime to Modulus iff Linear Congruence to 1 exists}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, m \in \Z$.
+The [[Definition:Linear Congruence|linear congruence]]:
+:$a x \equiv 1 \pmod m$
+has a solution $x$ {{iff}} $a$ and $m$ are [[Definition:Coprime Integers|coprime]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a x
+      | o = \equiv
+      | r = 1
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | lll= \iff
+      | ll= \exists y \in \Z:
+      | l = a x + m y
+      | r = 1
+      | c = Definition of [[Definition:Congruence Modulo Integer|congruence]]
+}}
+{{end-eqn}}
+From [[Integer Combination of Coprime Integers]]:
+: $a \perp m \iff \exists x, y \in \Z: a x + m y = 1$
+That is, such an $x$ exists {{iff}} $a$ and $m$ are [[Definition:Coprime Integers|coprime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Coprime to Modulus iff Linear Congruence to 1 exists/Corollary}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+The [[Definition:Linear Congruence|linear congruence]]:
+:$a x \equiv 1 \pmod p$
+has a solution $x$ {{iff}} $a \not \equiv 0 \pmod p$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Congruence (Number Theory)|congruence]]:
+:$a \not \equiv 0 \pmod p \iff p \nmid a$
+where $p \nmid a$ denotes that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $a$.
+From [[Prime not Divisor implies Coprime]]:
+:$p \nmid a \iff p \perp a$
+where $p \perp a$ denotes that $p$ and $a$ are [[Definition:Coprime Integers|coprime]].
+The result follows from [[Integer Coprime to Modulus iff Linear Congruence to 1 exists]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Linear Congruence/Existence}
+Tags: Solution of Linear Congruence
+
+\begin{theorem}
+$a x \equiv b \pmod n$ has at least one [[Definition:Polynomial Congruence/Solution|solution]] {{iff}}:
+: $\gcd \set {a, n} \divides b$
+that is, {{iff}} $\gcd \set {a, n}$ is a [[Definition:Divisor of Integer|divisor]] of $b$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$.
+Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$.
+Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of [[Definition:Congruence (Number Theory)|congruence]].
+Thus $x = x_0, y = y_0$ is a solution to the [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$.
+On the other hand, if $x = x_0, y = y_0$ is a solution to the [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$, then it follows that $a x \equiv b \pmod n$.
+Hence:
+: the problem of finding all integers satisfying the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$ 
+is the same problem as:
+: the problem of finding all the $x$ values in the [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$.
+From [[Solution of Linear Diophantine Equation]]:
+The [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$ has at least one [[Definition:Polynomial Congruence/Solution|solution]] {{iff}}:
+:$\gcd \set {a, n} \divides b$
+Hence the result.
+{{qed}}
+[[Category:Solution of Linear Congruence]]
+i2q79zyugpt7691hpi24jfc0wf67dz4
+\end{proof}<|endoftext|>
+\section{Solution of Linear Congruence/Number of Solutions}
+Tags: Solution of Linear Congruence
+
+\begin{theorem}
+Let $\gcd \set {a, n} = d$.
+Then $a x \equiv b \pmod n$ has $d$ solutions which are given by the [[Definition:Unique|unique]] solution modulo $\dfrac n d$ of the [[Definition:Linear Congruence|congruence]]:
+: $\dfrac a d x \equiv \dfrac b d \paren {\bmod \dfrac n d}$
+\end{theorem}
+
+\begin{proof}
+From [[Solution of Linear Congruence/Existence|Solution of Linear Congruence: Existence]]:
+:the problem of finding all integers satisfying the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$ 
+is the same problem as:
+:the problem of finding all the $x$ values in the [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$.
+From [[Integers Divided by GCD are Coprime]]:
+:$\gcd \set {\dfrac a d, \dfrac n d} = 1$
+So the {{RHS}} has a unique solution modulo $\dfrac n d$, say:
+:$x \equiv x_1 \paren {\bmod \dfrac n d}$
+So the integers $x$ which satisfy $a x \equiv b \pmod n$ are exactly those of the form $x = x_1 + k \dfrac n d$ for some $k \in \Z$.
+Consider the set of integers:
+: $\set {x_1, x_1 + \dfrac n d, x_1 + 2 \dfrac n d, \ldots, x_1 + \paren {d - 1} \dfrac n d}$
+None of these are congruent modulo $n$ and none differ by as much as $n$.
+Further, for any $k \in Z$, we have that $x_1 + k \dfrac n d$ is congruent modulo $n$ to one of them.
+To see this, write $k = d q + r$ where $0 \le r < d$ from the [[Division Theorem]].
+Then:
+{{begin-eqn}}
+{{eqn | l = x_1 + k \frac n d
+      | r = x_1 + \paren {d q + r} \frac n d
+      | c = 
+}}
+{{eqn | r = x_1 + n q + r \frac n d
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = x_1 + r \frac n d
+      | rr= \pmod n
+      | c = 
+}}
+{{end-eqn}}
+So these are the $d$ solutions of $a x \equiv b \pmod n$.
+{{qed}}
+[[Category:Solution of Linear Congruence]]
+npnsey8asyo4dq5g5ub0fbwfcrigw4m
+\end{proof}<|endoftext|>
+\section{Solution of Linear Congruence/Unique iff Coprime to Modulus}
+Tags: Solution of Linear Congruence
+
+\begin{theorem}
+If $\gcd \set {a, n} = 1$, then $a x \equiv b \pmod n$ has a [[Definition:Unique|unique]] solution.
+\end{theorem}
+
+\begin{proof}
+From [[Solution of Linear Congruence/Existence|Solution of Linear Congruence: Existence]]:
+:the problem of finding all integers satisfying the [[Definition:Linear Congruence|linear congruence]] $a x \equiv b \pmod n$ 
+is the same problem as:
+:the problem of finding all the $x$ values in the [[Definition:Linear Diophantine Equation|linear Diophantine equation]] $a x - n y = b$.
+Let:
+:$\gcd \set {a, n} = 1$
+Let $x = x_0, y = y_0$ be one solution to the [[Definition:Linear Diophantine Equation|linear Diophantine equation]]:
+:$a x - n y = b$
+From [[Solution of Linear Diophantine Equation]], the general solution is:
+:$\forall k \in \Z: x = x_0 + n k, y = y_0 + a k$
+But:
+:$\forall k \in \Z: x_0 + n k \equiv x_0 \pmod n$
+Hence $x \equiv x_0 \pmod n$ is the only solution of $a x \equiv b \pmod n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Direct Product of Modulo Integers is Isomorphic to Ring Modulo Product iff Coprime}
+Tags: Commutative Algebra, Modulo Arithmetic
+
+\begin{theorem}
+Let $m, n \in \Z_{>1}$.
+Let $\struct {\Z_m, +_m, \times_m}$ and $\struct {\Z_n, +_n, \times_n}$ be the [[Definition:Ring of Integers Modulo m|rings of integers modulo $m$ and $n$]] respectively.
+Let $\struct {\Z_m \times \Z_n}$ be the [[Definition:Ring Direct Product|direct product]] of $\Z_m$ and $\Z_n$.
+Let $\struct {\Z_{m n}, +_{m n}, \times_{m n} }$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $mn$]].
+Then $\struct {\Z_m \times \Z_n}$ is [[Definition:Ring Isomorphism|isomorphic]] to $\struct {\Z_{m n}, +_{m n}, \times_{m n} }$ {{iff}} $m$ and $n$ are [[Definition:Coprime Integers|coprime]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|First a homomorphism is established between $\struct {\Z_m \times \Z_n}$ and $\struct {\Z_m \times \Z_n}$. Then it is demonstrated that $\phi: \struct {\Z_m \times \Z_n} \to \Z_{m n}$ is a bijection. Some of this has already been done, I'm fairly sure.}}
+\end{proof}<|endoftext|>
+\section{Trivial Norm on Division Ring is Norm}
+Tags: Division Rings, Trivial Norms
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]], and denote its [[Definition:Ring Zero|ring zero]] by $0_R$.
+Then the [[Definition:Trivial Norm on Division Ring|trivial norm]] $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
+:$\norm x = \begin{cases}
+  0 & \text { if } x = 0_R\\
+  1 & \text { otherwise}
+\end{cases}$
+defines a [[Definition:Norm on Division Ring|norm]] on $R$.
+\end{theorem}
+
+\begin{proof}
+Proving each of the [[Definition:Norm on Division Ring|norm axioms]] one by one:
+=== Proving $(N1) : \forall x \in R: \norm x = 0 \iff x = 0_R$ ===
+This follows directly from the definition of the [[Definition:Trivial Norm on Division Ring|trivial norm]].
+{{qed|lemma}}
+=== Proving $(N2) : \forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$ ===
+If $x = 0_R$:
+{{begin-eqn}}
+{{eqn | l = \norm {x \circ y}
+      | r = \norm {0_R \circ y}
+      | c = by substitution
+}}
+{{eqn | r = \norm {0_R}
+      | c = [[Ring Product with Zero]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Trivial Norm on Division Ring}}
+}}
+{{eqn | r = 0 \times \norm y
+      | c = 
+}}
+{{eqn | r = \norm x \times \norm y
+      | c = since $x = 0_R$
+}}
+{{end-eqn}}
+The reasoning is similar if $y = 0_R$.
+If $x,y \ne 0_R$, then $x \circ y \ne 0_R$ by alternative definition $(3)$ of [[Definition:Division Ring|division rings]]. We get:
+{{begin-eqn}}
+{{eqn | l = \norm {x \circ y}
+      | r = 1
+      | c = since $x \circ y \ne 0_R$
+}}
+{{eqn | r = 1 \times 1
+      | c = 
+}}
+{{eqn | r = \norm x \times \norm y
+      | c = since $x, y \ne 0_R$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Proving $(N3) : \forall x, y \in R: \norm {x + y} \le \norm x + \norm y$ ===
+If $x = 0_R$:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | r = \norm {0_R + y}
+      | c = 
+}}
+{{eqn | r = \norm y
+      | c = 
+}}
+{{eqn | r = 0 + \norm y
+      | c = 
+}}
+{{eqn | r = \norm x + \norm y
+      | c = since $x = 0_R$
+}}
+{{end-eqn}}
+The reasoning is similar if $y = 0_R$.
+If $x, y \ne 0_R$:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | o = \le
+      | r = 1
+      | c = 
+}}
+{{eqn | o = <
+      | r = 1 + 1
+      | c = 
+}}
+{{eqn | r = \norm x + \norm y
+      | c = since $x, y \ne 0_R$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Division Rings]]
+[[Category:Trivial Norms]]
+maljxp1yu5tbbkzkk6jv4dfoj081df2
+\end{proof}<|endoftext|>
+\section{Trivial Norm on Division Ring is Non-Archimedean}
+Tags: Trivial Norms
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|ring zero]] is $0_R$.
+Then the [[Definition:Trivial Norm on Division Ring|trivial norm]] $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
+:$\norm x = \begin{cases}
+  0 & : x = 0_R \\
+  1 & : \text{ otherwise}
+\end{cases}$
+is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]]:
+:$\norm {x + y} \le \max \set {\norm x, \norm y}$
+\end{theorem}
+
+\begin{proof}
+Let $x, y = 0_R$.
+Then:
+:$\norm x, \norm y = 0$
+Therefore:
+:$\max \set {\norm x, \norm y} = 0$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | r = \norm {0_R + 0_R}
+      | c = 
+}}
+{{eqn | r = \norm {0_R}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{eqn | r = \max \set {\norm x, \norm y} 
+      | c = 
+}}
+{{end-eqn}}
+Let $x \ne 0_R$ or $y \ne 0_R$.
+Then:
+:$\norm x = 1$ or $\norm y = 1$
+Therefore:
+:$\max \set {\norm x, \norm y} = 1$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | o = \le
+      | r = 1
+      | c = {{Defof|Trivial Norm on Division Ring}}
+}}
+{{eqn | r = \max \set {\norm x, \norm y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Trivial Norms]]
+s5has14axjeaicsn7dkar1jr2vq0ibz
+\end{proof}<|endoftext|>
+\section{Ring of Integers Modulo m is Ring}
+Tags: Commutative Algebra, Ring of Integers Modulo m
+
+\begin{theorem}
+For all $m \in \N: m \ge 2$, the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]]:
+:$\struct {\Z_m, +_m, \times_m}$
+is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\eqclass 1 m$]].
+The [[Definition:Ring Zero|zero]] of $\struct {\Z_m, +_m, \times_m}$ is $\eqclass 0 m$.
+\end{theorem}
+
+\begin{proof}
+First we check the [[Definition:Ring Axioms|ring axioms]]:
+:$\text A$: The [[Integers Modulo m under Addition form Abelian Group |Integers Modulo $m$ under Addition form Abelian Group]].
+:From [[Modulo Addition has Identity]], $\eqclass 0 m$ is the [[Definition:Identity Element|identity]] of the [[Definition:Additive Group of Ring|additive group]] $\struct {\Z_m, +_m}$.
+From [[Integers Modulo m under Multiplication form Commutative Monoid]]:
+:$\text M 0$: $\struct {\Z_m, \times_m}$ is [[Definition:Closed Algebraic Structure|closed]].
+:$\text M $: $\struct {\Z_m, \times_m}$ is [[Definition:Associative|associative]].
+:$\text M 2$: $\struct {\Z_m, \times_m}$ has an [[Definition:Identity Element|identity]] $\eqclass 1 m$.
+:$\text C$: $\struct {\Z_m, \times_m}$ is [[Definition:Commutative Operation|commutative]].
+Then:
+:$\text D$: [[Modulo Multiplication Distributes over Modulo Addition|$\times_m$ distributes over $+_m$ in $\Z_m$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece}
+Tags: Piecewise Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]]:
+{{:Definition:Piecewise Continuous Function with One-Sided Limits}}
+Then:
+:for all $i \in \left\{ {1, 2, \ldots, n}\right\}$, $f$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ .
+\end{theorem}
+
+\begin{proof}
+We have that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{ {1, 2, \ldots, n}\right\}$.
+Since $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]], the [[Definition:One-Sided Limit of Real Function|one-sided limits]]:
+: $\displaystyle \lim_{x \mathop \to {x_{i−1} }^+} f \left({x}\right)$
+and:
+: $\displaystyle \lim_{x \mathop \to {x_i}^-} f \left({x}\right)$
+exist for every $i \in \left\{{1, 2, \ldots, n}\right\}$.
+Thus, from [[Extendability Theorem for Function Continuous on Open Interval]], a function $f_i$ exists that satisfies:
+: $f_i$ is defined on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$
+: $f_i$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$
+: $f_i$ equals $f$ on $\left({x_{i − 1} \,.\,.\, x_i}\right)$
+for an arbitrary $i$ in $\left\{{1, 2, \ldots, n}\right\}$
+From [[Continuous Function on Closed Interval is Uniformly Continuous]], $f_i$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$.
+We have that $\left({x_{i − 1} \,.\,.\, x_i}\right)$ is a [[Definition:Subset|subset]] of $\left[{x_{i − 1} \,.\,.\, x_i}\right]$.
+Hence $f_i$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.
+As $f_i$ equals $f$ on $\left({x_{i − 1} \,.\,.\, x_i}\right)$, $f$ also is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.
+Since $i$ is arbitrary, this results holds for all $i \in \left\{{1, 2, \ldots, n}\right\}$.
+{{qed}}
+[[Category:Piecewise Continuous Functions]]
+1o4dyhf6ae3ngxqoywpr90p0t6neql1
+\end{proof}<|endoftext|>
+\section{Principle of Definition by Structural Induction}
+Tags: Formal Systems
+
+\begin{theorem}
+Let $\LL$ be a [[Definition:Formal Language|formal language]].
+Let the [[Definition:Formal Grammar|formal grammar]] of $\LL$ be a [[Definition:Bottom-Up Grammar|bottom-up grammar]] with [[Definition:Unique Parsability|unique parsability]].
+A [[Definition:Definition|definition]] $\map D \phi$ (in the [[Definition:Metalanguage|metalanguage]] of $\LL$) for all [[Definition:Well-Formed Formula|well-formed formulas]] $\phi$ of $\LL$ is uniquely specified by:
+:$(1): \quad$ A definition $\map D a$ for each [[Definition:Letter|letter]] $a$ of $\LL$
+:$(2): \quad$ For each [[Definition:Rule of Formation|rule of formation]] for $\LL$, a definition $\map D \phi$ of the resultant [[Definition:Well-Formed Formula|WFF]] $\phi$ in terms of the constituent [[Definition:Well-Formed Formula|WFF]]s' definitions $\map D {\phi_1}, \ldots, \map D {\phi_n}$.
+\end{theorem}
+
+\begin{proof}
+We apply the [[Principle of Structural Induction]] on the following [[Definition:Statement|statement]] $\map P \phi$:
+:The definition is specified for $\phi$
+The given hypotheses verify the conditions for the [[Principle of Structural Induction]].
+It follows that $\map D \phi$ is specified for each [[Definition:Well-Formed Formula|WFF]] $\phi$.
+Moreover, by virtue of the [[Definition:Unique Parsability|unique parsability]], it is ensured that each [[Definition:Well-Formed Formula|WFF]] $\phi$ has a [[Definition:Unique|unique]] definition $\map D \phi$.
+{{qed}}
+[[Category:Formal Systems]]
+9trc12ncmpr7m2cvo2w8qth75d0tt7m
+\end{proof}<|endoftext|>
+\section{Principle of Structural Induction}
+Tags: Formal Systems
+
+\begin{theorem}
+Let $\LL$ be a [[Definition:Formal Language|formal language]].
+Let the [[Definition:Formal Grammar|formal grammar]] of $\LL$ be a [[Definition:Bottom-Up Grammar|bottom-up grammar]].
+Let $\map P \phi$ be a [[Definition:Statement|statement]] (in the [[Definition:Metalanguage|metalanguage]] of $\LL$) about [[Definition:Well-Formed Formula|well-formed formulas]] $\phi$ of $\LL$.
+Then $P$ is true for all [[Definition:Well-Formed Formula|WFFs]] of $\LL$ {{iff}} both:
+:$\map P a$ is true for all [[Definition:Letter|letters]] $a$ of $\LL$,
+and, for each [[Definition:Rule of Formation|rule of formation]] of $\LL$, if $\phi$ is a [[Definition:Well-Formed Formula|WFF]] resulting from WFFs $\phi_1, \ldots, \phi_n$ by applying that rule, then:
+:$\map P \phi$ is true only if $\map P {\phi_1}, \ldots, \map P {\phi_n}$ are all true.
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be a [[Definition:Well-Formed Formula|WFF]] of $\LL$.
+Then $\phi$ is the result of applying finitely many [[Definition:Rule of Formation|rules of formation]] of $\LL$.
+If we can show that the result of each [[Definition:Rule of Formation|rule of formation]] satisfies $P$, we will have finished.
+Suppose now that for a [[Definition:Rule of Formation|rule of formation]] $\mathbf R$, all preceding rules have produced [[Definition:Well-Formed Formula|WFFs]] satisfying $P$.
+By the [[Definition:Bottom-Up Grammar|bottom-up]] nature of the [[Definition:Formal Grammar|formal grammar]] of $\LL$, $\mathbf R$ either:
+:Introduces a [[Definition:Letter|letter]] as a [[Definition:Well-Formed Formula|WFF]]; or
+:Constructs a new [[Definition:Well-Formed Formula|WFF]] from previously constructed ones.
+The two given hypotheses precisely ensure that the [[Definition:Well-Formed Formula|WFF]] resulting from $\mathbf R$ must also satisfy $P$.
+For the first [[Definition:Rule of Formation|rule of formation]] applied, all preceding rules have [[Definition:Vacuous Truth|vacuously]] produced [[Definition:Well-Formed Formula|WFFs]] satisfying $P$.
+But now we see that any subsequent [[Definition:Rule of Formation|rule of formation]] will satisfy this premise.
+In particular, it applies to the final [[Definition:Rule of Formation|rule of formation]], and hence $\map P \phi$ is true.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Multiplication form Monoid}
+Tags: Integer Multiplication, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers under Multiplication form Semigroup]], $\struct {\Z, \times}$ is a [[Definition:Semigroup|semigroup]].
+From [[Integer Multiplication Identity is One]], $\struct {\Z, \times}$ has an [[Definition:Identity Element|identity element]], which is $1$.
+All the criteria for being a [[Definition:Monoid|monoid]] are thus seen to be fulfilled.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Multiplication do not form Group}
+Tags: Integer Multiplication
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ does not form a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+In order to be classified as a [[Definition:Group|group]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, \times}$ needs to fulfil the [[Definition:Group Axioms|group axioms]].
+From [[Integers under Multiplication form Monoid]], $\struct {\Z, \times}$ forms a [[Definition:Monoid|monoid]].
+Therefore [[Definition:Group Axioms|group axioms]] $\text G 0$, $\text G 1$ and $\text G 2$ are satisfied.
+However, from [[Invertible Integers under Multiplication]], the only [[Definition:Integer|integers]] with [[Definition:Inverse Element|inverses]] under [[Definition:Integer Multiplication|multiplication]] are $1$ and $-1$.
+As not all [[Definition:Integer|integers]] have [[Definition:Inverse Element|inverses]], it follows that $\struct {\Z, \times}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Multiplication form Semigroup}
+Tags: Integer Multiplication, Examples of Semigroups
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+=== Closure ===
+[[Integer Multiplication is Closed]].
+{{qed|lemma}}
+=== Associativity ===
+[[Integer Multiplication is Associative]].
+{{qed|lemma}}
+The criteria for $\struct {\Z, \times}$ to be a [[Definition:Semigroup|semigroup]] are seen to be satisfied. 
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Bottom-Up Form of Top-Down Grammar defines same Formal Language}
+Tags: Formal Systems
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Formal Language|formal language]].
+Let $\mathcal T$ be a [[Definition:Top-Down Grammar|top-down grammar]] for $\mathcal L$.
+Let $\mathcal B$ be the [[Definition:Bottom-Up Form of Top-Down Grammar|bottom-up form]] of $\mathcal T$.
+Then $\mathcal B$ is also a [[Definition:Formal Grammar|formal grammar]] for $\mathcal L$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+If $\phi$ is a [[Definition:Letter|letter]], then it is clearly a $\mathcal T$-[[Definition:Well-Formed Formula|WFF]].
+For, it may be formed by replacing the starting [[Definition:Metasymbol|metasymbol]] of $\mathcal T$ by $\phi$.
+Suppose that $\phi$ is formed from [[Definition:Well-Formed Formula|WFFs]] $\phi_1, \ldots, \phi_n$ by the [[Definition:Rule of Formation|rule of formation]] $\mathbf R_{\mathcal B}$ of $\mathcal B$.
+Suppose also that each of $\phi_1, \ldots, \phi_n$ is also a $\mathcal T$-[[Definition:Well-Formed Formula|WFF]].
+Then by applying the [[Definition:Rule of Formation|rule of formation]] $\mathbf R$ of $\mathcal T$, we obtain a [[Definition:Collation|collation]] with [[Definition:Metasymbol|metasymbols]] $\psi_1, \ldots, \psi_n$.
+By assumption, we can apply [[Definition:Rule of Formation|rules of formation]] of $\mathcal T$ to each $\psi_i$ to yield the corresponding $\mathcal T$-[[Definition:Well-Formed Formula|WFF]] $\phi_i$.
+It follows that $\phi$ is also a $\mathcal T$-[[Definition:Well-Formed Formula|WFF]].
+By the [[Principle of Structural Induction]], each $\mathcal B$-[[Definition:Well-Formed Formula|WFF]] is also a $\mathcal T$-[[Definition:Well-Formed Formula|WFF]].
+Conversely, suppose that $\phi$ is a $\mathcal T$-[[Definition:Well-Formed Formula|WFF]].
+Let it be formed by applying the [[Definition:Rule of Formation|rules of formation]] $\mathbf R_1, \ldots, \mathbf R_n$, in succession.
+We will prove that each [[Definition:Metasymbol|metasymbol]] in the inputs of these [[Definition:Rule of Formation|rules of formation]] will be replaced by a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+In particular, then, $\phi$ will be a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+Since $\mathbf R_n$ is the last [[Definition:Rule of Formation|rule of formation]] applied, it must replace a [[Definition:Metasymbol|metasymbol]] by a [[Definition:Letter|letter]].
+Since [[Definition:Letter|letters]] are $\mathcal B$-[[Definition:Well-Formed Formula|WFFs]], $\mathbf R_n$ satisfies the assertion.
+Suppose that we have established that all [[Definition:Metasymbol|metasymbols]] in the result of $\mathbf R_i$ will be replaced by $\mathcal B$-[[Definition:Well-Formed Formula|WFFs]].
+Then also the [[Definition:Metasymbol|metasymbol]] $\psi$ that $\mathbf R_i$ replaces by a new [[Definition:Collation|collation]] will be replaced by a [[Definition:Well-Formed Formula|WFF]].
+This is because the [[Definition:Collation|collation]] resulting from replacing $\psi$ according to $\mathbf R_i$ contains [[Definition:Metasymbol|metasymbols]] $\psi_i$, which by assumption will be replaced by corresponding $\mathcal B$-[[Definition:Well-Formed Formula|WFFs]] $\phi_i$.
+Now the [[Definition:Rule of Formation|rule of formation]] $\left({\mathbf R_i}\right)_{\mathcal B}$ ensures that $\psi$ will be replaced by a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+So each [[Definition:Metasymbol|metasymbol]] $\psi$ in the input of $\mathbf R_i$ will be replaced by a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+Hence $\phi$ is a $\mathcal B$-[[Definition:Well-Formed Formula|WFF]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Characteristic of Ring}
+Tags: Rings with Unity
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+{{TFAE|def = Characteristic of Ring}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 is equivalent to Definition 3 ===
+By definition of [[Definition:Order of Group Element|order]], the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the [[Definition:Identity of Group|identity]] of the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$ of $\struct {R, +, \circ}$.
+Hence if the order of $1_R$ is finite, the definitions coincide.
+If the order of $1_R$ is infinite, there is no such $p$.
+Then both definitions give $\Char R = 0$, so they coincide as well.
+{{qed|lemma}}
+=== Definition 1 is equivalent to Definition 2 ===
+Let $g: \Z \to R$ be the [[Definition:Initial Homomorphism from Integers to Ring with Unity|initial homomorphism]], with $\map g n = n \cdot 1_R$.
+By [[Kernel of Ring Homomorphism is Ideal]], $\ker g$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+By [[Ring of Integers is Principal Ideal Domain]], there exists a [[Definition:Unique|unique]] $p \in \Z_{\ge 0}$ such that $\ker g$ is the [[Definition:Principal Ideal of Ring|principal ideal]] $\ideal p$.
+Suppose there is a smallest $m$ such that $m \cdot 1_R = 0_R$.
+Then for any $k \in \Z$:
+:$\map g {k m} = k m \cdot 1_R = k \cdot 0_R = 0_R$.
+This gives $\ideal m \subseteq \ker g = \ideal p$.
+Let $a \in \ker g$.
+Then by the [[Division Theorem]]:
+:$\exists q, r \in \Z, 0 \le r < m: a = m q + r$
+As $a, m \in \ker g = \ideal p$ so does $r = a - m q$.
+Since $m$ is the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $m \cdot 1_R = 0_R$:
+:$\forall n \in \Z: 0 < n < m \implies \map g n = n \cdot 1_R \ne 0_R$
+This forces $r = 0$.
+Thus $a = m q \in \ideal m$.
+Hence we have $\ideal m = \ideal p$, giving $p = m$.
+Now consider the case where $m$ does not exist.
+Then for any $k \ne 0$:
+:$\map g k = k \cdot 1_R \ne 0_R$
+And we have:
+:$\map g 0 = 0 \cdot 1_R = 0_R$
+giving $\ker g = \set 0 = \ideal 0$.
+Both definitions give $\Char R = 0$.
+Therefore the definitions are equivalent in each case.
+{{qed}}
+[[Category:Rings with Unity]]
+7zcpiooxjpqxzrlej6cbfkafzp4apny
+\end{proof}<|endoftext|>
+\section{Canonical Form of Rational Number is Unique}
+Tags: Rational Numbers
+
+\begin{theorem}
+The [[Definition:Canonical Form of Rational Number|canonical form]] of a [[Definition:Rational Number|rational number]] is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let $r \in \Q$ be a [[Definition:Rational Number|rational number]].
+Let $\dfrac p q$ and $\dfrac {p'} {q'}$ be two [[Definition:Canonical Form of Rational Number|canonical forms]] of $r$.
+{{WLOG}}, assume $q \le q'$.
+By [[Equality of Rational Numbers]]:
+:$p q' = p' q$
+Therefore, $q'$ [[Definition:Divisor of Integer|divides]] $p' q$.
+As $\dfrac {p'} {q'}$ is a [[Definition:Canonical Form of Rational Number|canonical form]] of $r$, by definition $q'$ is [[Definition:Coprime Integers|coprime]] to $p'$.
+By [[Euclid's Lemma]] $q'$ [[Definition:Divisor of Integer|divides]] $q$.
+$q$ and $q'$ are both [[Definition:Positive Integer|positive]].
+So from [[Absolute Value of Integer is not less than Divisors/Corollary|Absolute Value of Integer is not less than Divisors: Corollary]]:
+:$q' \le q$
+As  $q' \le q$ and $q \le q'$ it follows that $q = q'$.
+It then follows that $p = p'$.
+Hence $\dfrac p q$ and $\dfrac {p'} {q'}$ are the same [[Definition:Unique|unique]] [[Definition:Canonical Form of Rational Number|canonical form]] of $r$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Canonical Form of Rational Number}
+Tags: Rational Numbers
+
+\begin{theorem}
+Let $r \in \Q$.
+Then:
+:$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$
+That is, every [[Definition:Rational Number|rational number]] can be expressed in its [[Definition:Canonical Form of Rational Number|canonical form]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Rational Number|set of rational numbers]] is the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Set|set]] of [[Definition:Integer|integers]].
+From [[Divided by Positive Element of Field of Quotients]]:
+:$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$
+Now if $s \perp t$, our task is complete.
+Otherwise, let:
+:$\gcd \set {s, t} = d$
+where $\gcd \set {s, t}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $s$ and $t$.
+Let $s = p d, t = q d$.
+We have that $t, d \in \Z_{>0}$
+Therefore $q \in \Z_{>0}$ also.
+From [[Integers Divided by GCD are Coprime]]:
+:$p \perp q$
+Also:
+{{begin-eqn}}
+{{eqn | l = \frac s t
+      | r = \frac {p d} {q d}
+      | c = 
+}}
+{{eqn | r = \frac p q \frac d d
+      | c = 
+}}
+{{eqn | r = \frac p q 1
+      | c = 
+}}
+{{eqn | r = \frac p q
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$r = \dfrac p q$
+where $p \perp q$ and $q \in \Z_{>0}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Dyadic Rational between two Rationals}
+Tags: Rational Numbers
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Rational Number|rational numbers]] such that $a < b$.
+Then there exist [[Definition:Integer|integers]] $m$ and $r$ such that:
+:$a < \dfrac m {2^r} < b$
+That is, there exists a [[Definition:Dyadic Rational|dyadic rational]] between any pair of [[Definition:Rational Number|rational numbers]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|An exercise for the student.}}
+\end{proof}<|endoftext|>
+\section{Field with 4 Elements has only Order 2 Elements}
+Tags: Galois Fields, Field with 4 Elements has only Order 2 Elements
+
+\begin{theorem}
+Let $\struct {\GF, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] which has exactly $4$ [[Definition:Element|elements]].
+Then:
+:$\forall a \in \GF: a + a = 0_\GF$
+where $0_\GF$ is the [[Definition:Field Zero|zero]] of $\GF$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Field (Abstract Algebra)|field]], both the [[Definition:Algebraic Structure|algebraic structures]] $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are [[Definition:Abelian Group|(abelian) groups]], where $\GF^* := \GF \setminus \set 0$.
+By definition:
+:$\struct {\GF, +}$ is of [[Definition:Order of Structure|order $4$]]
+:$\struct {\GF^*, \times}$ is of [[Definition:Order of Structure|order $3$]].
+From [[Classification of Groups of Order up to 15]], there are only two possibilities:
+:$(1): \quad \struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$
+:$(2): \quad \struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$.
+In case $(2)$:
+:$\forall a \in \GF: a + a = 0_\GF$
+but in case $(1)$:
+:$\exists a, b \in \GF: a + a = b \ne 0_\GF$ where $b + b = 0_\GF$
+as there exists an element of $\struct {\GF, +}$ of [[Definition:Order of Group Element|order $4$]].
+{{AimForCont}} $(1)$ describes a [[Definition:Field (Abstract Algebra)|field]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {b + b} \times b
+      | r = 0_\GF \times b
+      | c = as $b + b = 0_\GF$ from above
+}}
+{{eqn | r = 0_\GF
+      | c = as $0_\GF$ is the [[Definition:Field Zero|zero]] of $\struct {\GF, +, \times}$
+}}
+{{end-eqn}}
+But then:
+{{begin-eqn}}
+{{eqn | l = \paren {b + b} \times b
+      | r = \paren {b \times b} + \paren {b \times b}
+      | c = as $\times$ is [[Definition:Distributive Operation|distributive]] over $+$
+}}
+{{end-eqn}}
+As $\struct {\GF, +} \cong \Z_4$, both $a + a = b$ and $\paren {-a} + \paren {-a} = b$.
+The only way for $\paren {b \times b} + \paren {b \times b} = 0_\GF$ is for $b \times b = b$ or $b \times b = 0_\GF$.
+The second case is eliminated as $0_\GF \notin \struct {\GF^*, \times}$.
+So it must be the case that $b \times b = b$.
+So as $\struct {\GF^*, \times} \cong \Z_3$ it must follow that $a \times b = a, a \times \paren {-a} = b$ and then $a \times a = -a$.
+It follows that the [[Definition:Cayley Table|Cayley tables]] of $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ must therefore be as follows:
+:$\begin{array}{c|cccc}
++ & 0_\GF & a & b & -a \\
+\hline
+0_\GF & 0_\GF & a & b & -a \\
+a & a & b & -a & 0_\GF \\
+b & b & -a & 0_\GF & a \\
+-a & -a & 0_\GF & a & b \\
+\end{array}
+\qquad
+\begin{array}{c|cccc}
+\times & 0_\GF & b & a & -a \\
+\hline
+0_\GF & 0_\GF & 0_\GF & 0_\GF & 0_\GF \\
+b & 0_\GF & b & a & -a \\
+a & 0_\GF & a & -a & b \\
+-a & 0_\GF & -a & b & a \\
+\end{array}$
+{{begin-eqn}}
+{{eqn | l = \paren {a + b} \times a
+      | r = \paren {-a} \times a
+      | c = as $a + b = -a$ from above
+}}
+{{eqn | r = b
+      | c = as $\paren {-a} \times a = b$ from above
+}}
+{{end-eqn}}
+But:
+{{begin-eqn}}
+{{eqn | l = \paren {a \times a} + \paren {b \times a}
+      | r = \paren {-a} + a
+      | c = from above
+}}
+{{eqn | r = 0_\GF
+      | c = from above
+}}
+{{end-eqn}}
+So:
+:$\paren {a + b} \times a \ne \paren {a \times a} + \paren {b \times a}$
+demonstrating that $\times$ is not [[Definition:Distributive Operation|distributive]] over $+$.
+Thus $\GF$ as has been defined:
+:$\struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$
+is not a [[Definition:Field (Abstract Algebra)|field]].
+It follows that our supposition that $\struct {\GF, +} \cong \Z_4$ was false.
+Thus, if $\GF$ is a [[Definition:Field (Abstract Algebra)|field]], then $\struct {\GF, +} \cong \Z_2 \times \Z_2$.
+That is:
+:$\forall a \in \GF: a + a = 0_\GF$
+as needed to be proved.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\struct {\GF, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Order of Structure|order $4$]] whose [[Definition:Field Zero|zero]] is $0_\GF$.
+By definition, $\GF$ is a [[Definition:Galois Field|Galois field]].
+The [[Definition:Additive Group of Field|additive group]] $\struct {\GF, +}$ of $\GF$ can be one of two:
+:$(1): \quad$ the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $4$]], [[Definition:Generator of Cyclic Group|generated]] by the [[Definition:Identity Element|identity]] of $\struct {\GF, +}$ which is $0_\GF$
+or:
+:$(2): \quad$ the [[Definition:Klein Four-Group|Klein $4$-group]], whose [[Definition:Element|elements]] are all of the form $a + a = 0_\GF$.
+{{AimForCont}} $(1)$ is the [[Definition:Additive Group of Field|additive group]] of a [[Definition:Field (Abstract Algebra)|field]].
+Then the [[Definition:Characteristic of Field|characteristic]] of $\GF$ is $4$.
+But from [[Characteristic of Galois Field is Prime]] it follows that $\GF$ is not a [[Definition:Field (Abstract Algebra)|field]].
+From that [[Definition:Contradiction|contradiction]] it follows that the [[Definition:Additive Group of Field|additive group]] $\struct {\GF, +}$ of $F$ cannot be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $4$]].
+Hence for $\GF$ to be a [[Definition:Field (Abstract Algebra)|field]] at all, it is [[Definition:Necessary Condition|necessary]] for $\struct {\GF, +}$ to be the [[Definition:Klein Four-Group|Klein $4$-group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Count of Truth Functions}
+Tags: Combinatorics, Truth Functions
+
+\begin{theorem}
+There are $2^{\paren {2^k} }$ [[Definition:Distinct Elements|distinct]] [[Definition:Truth Function|truth functions]] on $k$ variables.
+\end{theorem}
+
+\begin{proof}
+Let $f: \mathbb B^k \to \mathbb B$ be a [[Definition:Truth Function|truth function]].
+The [[Definition:Domain of Mapping|domain]] of $f$ has $2^k$ elements, from [[Cardinality of Cartesian Product]].
+The result follows from [[Cardinality of Set of All Mappings]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unary Truth Functions}
+Tags: Truth Functions
+
+\begin{theorem}
+There are $4$ distinct [[Definition:Unary|unary]] [[Definition:Truth Function|truth functions]]:
+:$(1): \quad$ The [[Definition:Constant Mapping|constant function]] $\map f p = \F$
+:$(2): \quad$ The [[Definition:Constant Mapping|constant function]] $\map f p = \T$
+:$(3): \quad$ The [[Definition:Identity Mapping|identity function]] $\map f p = p$
+:$(4): \quad$ The [[Definition:Logical Not|logical not]] function $\map f p = \neg p$
+\end{theorem}
+
+\begin{proof}
+From [[Count of Truth Functions]] there are $2^{\paren {2^1} } = 4$ distinct [[Definition:Truth Function|truth functions]] on $1$ variable.
+These can be depicted in a [[Definition:Truth Table|truth table]] as follows:
+:$\begin{array}{|c|cccc|} \hline
+p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
+\hline
+\T & \T & \T & \F & \F \\
+\F & \T & \F & \T & \F \\
+\hline
+\end{array}$
+$\circ_1$: Whether $p = \T$ or $p = \F$, $\map {\circ_1} p = \T$.
+Thus $\circ_1$ is the [[Definition:Constant Mapping|constant function]] $\map {\circ_1} p = \T$.
+$\circ_2$: We have:
+:$(1): \quad p = \T \implies \map {\circ_2} p = \T$
+:$(2): \quad p = \F \implies \map {\circ_2} p = \F$
+Thus $\circ_2$ is the [[Definition:Identity Mapping|identity function]] $\map {\circ_2} p = p$.
+$\circ_3$: We have:
+:$(1): \quad p = \T \implies \map {\circ_3} p = \F$
+:$(2): \quad p = \F \implies \map {\circ_3} p = \T$
+Thus $\circ_3$ is the [[Definition:Logical Not|logical not]] function $\map {\circ_3} p = \neg p$.
+$\circ_4$: Whether $p = \T$ or $p = \F$, $\map {\circ_4} p = \F$.
+Thus $\circ_4$ is the [[Definition:Constant Mapping|constant function]] $\map {\circ_4} p = \F$.
+All four have been examined, and there are no other [[Definition:Unary|unary]] [[Definition:Truth Function|truth functions]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binary Truth Functions}
+Tags: Truth Functions
+
+\begin{theorem}
+There are $16$ [[Definition:Distinct Elements|distinct]] [[Definition:Binary Operation|binary]] [[Definition:Truth Function|truth functions]]:
+* Two [[Definition:Constant Mapping|constant functions]]:
+** $\map {f_\F} {p, q} = \F$
+** $\map {f_\T} {p, q} = \T$
+* Two [[Definition:Projection (Mapping Theory)|projections]]:
+** $\map {\pr_1} {p, q} = p$
+** $\map {\pr_2} {p, q} = q$
+* Two [[Definition:Logical Not|negated]] [[Definition:Projection (Mapping Theory)|projections]]:
+** $\map {\overline {\pr_1} } {p, q} = \neg p$
+** $\map {\overline {\pr_2} } {p, q} = \neg q$
+* The [[Definition:Conjunction|conjunction]]: $p \land q$
+* The [[Definition:Disjunction|disjunction]]: $p \lor q$
+* Two [[Definition:Conditional|conditionals]]:
+** $p \implies q$
+** $q \implies p$
+* The [[Definition:Biconditional|biconditional]]: $p \iff q$
+* The [[Definition:Exclusive Or|exclusive or]]: $\map \neg {p \iff q}$
+* Two [[Definition:Logical Not|negated]] [[Definition:Conditional|conditionals]]:
+** $\map \neg {p \implies q}$
+** $\map \neg {q \implies p}$
+* The [[Definition:Logical NAND|NAND]]: $p \uparrow q$
+* The [[Definition:Logical NOR|NOR]]: $p \downarrow q$
+\end{theorem}
+
+\begin{proof}
+From [[Count of Truth Functions]] there are $2^{\paren {2^2} } = 16$ distinct [[Definition:Truth Function|truth functions]] on $2$ variables.
+These can be depicted in a [[Definition:Truth Table|truth table]] as follows:
+$\begin{array}{|r|cccc|} \hline
+p                                & \T & \T & \F & \F \\
+q                                & \T & \F & \T & \F \\
+\hline
+\map {f_\T} {p, q}               & \T & \T & \T & \T \\
+p \lor q                         & \T & \T & \T & \F \\
+p \impliedby q                   & \T & \T & \F & \T \\
+\map {\pr_1} {p, q}              & \T & \T & \F & \F \\
+p \implies q                     & \T & \F & \T & \T \\
+\map {\pr_2} {p, q}              & \T & \F & \T & \F \\
+p \iff q                         & \T & \F & \F & \T \\
+p \land q                        & \T & \F & \F & \F \\
+p \uparrow q                     & \F & \T & \T & \T \\
+\map \neg {p \iff q}             & \F & \T & \T & \F \\
+\map {\overline {\pr_2} } {p, q} & \F & \T & \F & \T \\
+\map \neg {p \implies q}         & \F & \T & \F & \F \\
+\map {\overline {\pr_1} } {p, q} & \F & \F & \T & \T \\
+\map \neg {p \impliedby q}       & \F & \F & \T & \F \\
+p \downarrow q                   & \F & \F & \F & \T \\
+\map {f_\F} {p, q}               & \F & \F & \F & \F \\
+\hline
+\end{array}$
+That accounts for all $16$ of them.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition}
+Tags: Prime Numbers, Rational Addition, Algebraic Closure
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $A_p$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] which, when expressed in [[Definition:Canonical Form of Rational Number|canonical form]] has a [[Definition:Numerator|numerator]] which is [[Definition:Divisor of Integer|divisible]] by $p$.
+Then $A_p$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Addition|rational addition]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in A_p$.
+Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
+:$n_1, n_2 \in \Z$
+:$d_1, d_2 \in \Z_{>0}$
+:$p n_1 \perp d_1, p n_2 \perp d_2$
+Then:
+{{begin-eqn}}
+{{eqn | l = a + b
+      | r = \frac {p n_1} {d_1} + \frac {p n_2} {d_2}
+      | c = 
+}}
+{{eqn | r = \frac {p n_1 d_2 + p n_2 d_1} {d_1 d_2}
+      | c = {{Defof|Rational Addition}}
+}}
+{{eqn | r = \frac {p \left({n_1 d_2 + n_2 d_1}\right)} {d_1 d_2}
+      | c = 
+}}
+{{end-eqn}}
+From [[Euclid's Lemma for Prime Divisors]], if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.
+But neither of these is the case, so $p \nmid d_1 d_2$.
+Hence by [[Prime not Divisor implies Coprime]]:
+:$p \perp d_1 d_2$
+where $\perp$ denotes [[Definition:Coprime Integers|coprimeness]].
+So when $\dfrac {p \left({n_1 d_2 + n_2 d_1}\right)} {d_1 d_2}$ is expressed in [[Definition:Canonical Form of Rational Number|canonical form]], $p$ will still be a [[Definition:Divisor of Integer|divisor]] of the [[Definition:Numerator|numerator]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Hyperbolic Sine Logarithmic Formulation}
+Tags: Inverse Hyperbolic Sine
+
+\begin{theorem}
+{{finish|The true story is more complicated than this. The [[Definition:Inverse Hyperbolic Sine|inverse hyperbolic sine]] is a multifunction. So is $\ln$ in the complex plane. And $\sqrt {z^2 + 1}$ has two values, pos and neg, so you need to justify which one is taken. The best context to put this page is probably into [[Definition:Inverse Hyperbolic Sine/Arcsine]], where an explanation needs to be made as to why the principal value is taken thus.}}
+For any [[Definition:Complex Number|complex number]] $z \in \C$:
+:$\operatorname {arsinh} z = \map \ln {z + \sqrt {z^2 + 1} }$
+where $\operatorname {arsinh} z$ is the [[Definition:Inverse Hyperbolic Sine|inverse hyperbolic sine]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \sinh \operatorname {arsinh} z
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = z
+      | r = \frac {e^{\operatorname{arsinh} z} - e^{-\operatorname{arsinh} z} } 2
+      | c = {{Defof|Inverse Hyperbolic Sine}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 z e^{\operatorname{arsinh} z}
+      | r = e^{2 \operatorname{arsinh} z} - 1
+      | c = Multiplication by $2 e^{\operatorname{arsinh} z}$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 0
+      | r = e^{2 \operatorname{arsinh} z} - 2 z e^{\operatorname{arsinh} z} - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{\operatorname{arsinh} z}
+      | r = z + \sqrt {z^2 + 1}
+      | c = [[Quadratic Formula]], $e^z > 0, \sqrt {z^2 + 1} > z$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \operatorname{arsinh} z
+      | r = \map \ln {z + \sqrt {z^2 + 1 } }
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Inverse Hyperbolic Sine]]
+mkdlysytrenv0xs7a0zsgl31zghtzvk
+\end{proof}<|endoftext|>
+\section{Value of Cauchy Determinant}
+Tags: Cauchy Matrix, Determinants, Value of Cauchy Determinant
+
+\begin{theorem}
+Let $D_n$ be a [[Definition:Cauchy Determinant|Cauchy determinant of order $n$]]:
+:$\begin{vmatrix}
+\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
+\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\
+\end{vmatrix}$
+Then the value of $D_n$ is given by:
+:$D_n = \dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right) \left({y_j - y_i}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i + y_j}\right)}$
+If $D_n$ is given by:
+:$\begin{vmatrix}
+\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
+\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
+\end{vmatrix}$
+then its [[Definition:Determinant of Matrix|determinant]] is given by:
+:$D_n = \dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right) \left({y_i - y_j}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i - y_j}\right)}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = C
+      | r = \begin{bmatrix}
+\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
+\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
+\end{bmatrix}
+}}
+{{end-eqn}}
+To be proved:
+{{begin-eqn}}
+{{eqn | l = \det \paren {C} 
+      | r = \dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right) \left({y_i - y_j}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i - y_j}\right)}
+      | c = Knuth (1997) replacing $y_k \to -y_k$ in $C$ and $\det \paren {C}$
+}}
+{{end-eqn}}
+Assume hereafter that set $\set {x_1,\ldots,x_n,y_1,\ldots,y_n}$ consists of distinct values, because otherwise $\det \paren {C}$ is undefined or zero.
+'''Preliminaries''':
+[[Vandermonde Matrix Identity for Cauchy Matrix]] supplies matrix equation
+:$\displaystyle (1)\quad - C = PV_x^{-1} V_y Q^{-1}$
+:Definitions of symbols:
+::$\displaystyle V_x=\paren {\begin{smallmatrix}
+1         & 1         & \cdots & 1 \\
+x_1       & x_2       & \cdots & x_n \\
+\vdots    & \vdots    & \ddots & \vdots \\
+x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\
+\end{smallmatrix} },\quad 
+V_y=\paren {\begin{smallmatrix}
+1         & 1         & \cdots & 1 \\
+y_1       & y_2       & \cdots & y_n \\
+\vdots    & \vdots    & \ddots & \vdots \\
+y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\
+\end{smallmatrix} }$ [[Definition:Vandermonde Matrix|Vandermonde matrices]]
+::$\displaystyle P= \paren {\begin{smallmatrix}
+p_1(x_1) &  \cdots & 0 \\
+\vdots   & \ddots  & \vdots \\
+0        & \cdots  & p_n(x_n) \\
+\end{smallmatrix} }, \quad
+Q= \paren {\begin{smallmatrix}
+p(y_1)  & \cdots  & 0 \\
+\vdots  & \ddots  & \vdots \\
+0       & \cdots  & p(y_n) \\
+\end{smallmatrix} }$  [[Definition:Diagonal Matrix|Diagonal matrices]]
+::$\displaystyle 
+p(x) = \prod_{i \mathop = 1}^n \paren {x - x_i},
+\quad
+\displaystyle 
+p_k(x) = \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i},
+\quad
+1 \mathop \le k \mathop \le n$ [[Definition:Polynomial/Complex Numbers|Polynomials]]
+'''Determinant of $C$ Calculation''':
+{{begin-eqn}}
+{{eqn | l = C
+      | r = \paren {-I}\,P\,V_x^{-1}\, V_y\, Q^{-1}
+      | c = [[Vandermonde Matrix Identity for Cauchy Matrix]] 
+Symbol $I$ is the $n\times n$ identity matrix.
+}}
+{{eqn | l = (2)\quad  \det \paren {C} 
+       | r = \det \paren { -I } \det \paren {P} \det \paren { V_x^{-1} } \det \paren { V_y } \det \paren { Q^{-1} }
+       | c = [[Determinant of Matrix Product]]
+}}
+{{eqn | r =  
+\paren { -1 }^n 
+\det \paren { I }
+\det \paren {P} \det \paren { V_x^{-1} } \det \paren { V_y } \det \paren { Q^{-1} }
+     | c = [[Effect of Elementary Row Operations on Determinant]] 
+Factor constant $-1$ from all rows of $-I$. 
+}}
+{{eqn | r =  
+\paren { -1 }^n 
+\det \paren { I }
+\dfrac
+{ \det \paren {P}  \det \paren { V_y } }
+{  \det \paren { Q } \det \paren { V_x } }
+      | c = [[Matrix Product with Adjugate Matrix]] and
+[[Determinant of Matrix Product]]
+}}
+{{end-eqn}}
+'''Lemma''': $\det \paren {P} = \paren {-1}^m \paren { \det \paren {V_x} }^2$ where $m=\frac 1 2 n \paren {n-1}$
+:'''Details''': Determinant $\det \paren {P}$ expands to:
+{{begin-eqn}}
+{{eqn | l =
+\,\,\prod_{j \mathop = 1}^n \map {p_j} {x_j} 
+      | r = 
+\prod_{j \mathop = 1}^n
+\,\, 
+\prod_{k \mathop = 1,\, k \mathop \neq j}^n \paren { x_j - x_k }
+      | c = Definition of polynomials $\map {p_j} x$.
+}}
+{{end-eqn}}
+:Pair factors $\paren {x_r - x_s}$ and  $\paren {x_s - x_r}$ into factor $- \paren {x_s - x_r}^2$, then:
+{{begin-eqn}}
+{{eqn | l = \,\,\det \paren {P} 
+      | r =  
+\paren {-1}^m 
+\, 
+\paren {
+\prod_{1 \mathop \leq r \mathop \lt s \mathop \le n} \paren { x_s - x_r }^2
+}
+      | c = where $m = 1 + \cdots + \paren {n-1} = \frac 1 2 n \paren {n-1}$
+}}
+{{eqn | r =
+\paren {-1}^m 
+\, 
+\paren { \det \paren {V_x} }^2
+      | c = [[Vandermonde Determinant]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Apply the '''Lemma''' to equation (2):
+{{begin-eqn}}
+{{eqn | l =
+\det \paren {C} 
+      | r =  
+\paren { -1 }^{n+m} 
+\,
+\paren {
+\dfrac
+{ \det \paren {V_x}  \det \paren { V_y } }
+{  \det \paren { Q } }
+}
+      | c = $m = \frac 1 2 n \paren {n-1}$
+}}
+{{eqn | r =  
+\paren { -1 }^{n+m}  
+\,\,
+\dfrac
+{ \displaystyle
+  \prod_{1 \mathop \le m \mathop \lt k \mathop \le n}^{\phantom n} \paren { x_k - x_m  }
+  \,
+  \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren { y_k - y_m  } 
+}
+{ \displaystyle
+    \prod_{k \mathop = 1}^n \map {p} {y_k} 
+}
+    | c = [[Vandermonde Determinant]] and
+[[Determinant of Diagonal Matrix]]
+}}
+{{eqn | r =
+\paren { -1 }^{n+m}  
+\,\,
+\dfrac
+{ \displaystyle
+  \prod_{1 \mathop \le m \mathop \lt k \mathop \le n}^{\phantom n} \paren { x_k - x_m  } \paren { y_k - y_m  } 
+}
+{ \displaystyle
+    \prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren { y_k - x_j }
+}
+    | c = Definition of $\map p x$.
+}}
+{{eqn | r =  
+\paren { -1 }^{n+m}  
+\,\,
+\dfrac
+{ \displaystyle
+  \paren {-1}^{m}\prod_{1 \mathop \le m \mathop \lt k \mathop \le n}^{\phantom n} \paren { x_k - x_m  } \paren { y_m - y_k } 
+}
+{ \displaystyle
+    \paren {-1}^{n^2}\,\,\prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren { x_j - y_k }
+}
+    | c = Factor out changed signs.
+}}
+{{eqn | r = 
+\left.
+\dfrac
+{ \displaystyle
+  \prod_{1 \mathop \le m \mathop \lt k \mathop \le n}^{\phantom n} \paren { x_k - x_m  } \paren { y_m - y_k } 
+}
+{ \displaystyle
+    \prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren { x_j - y_k }
+}
+\right.
+    | c = All signs cancel.
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Take the version of the [[Definition:Cauchy Matrix|Cauchy matrix]] defined such that $a_{ij} = \dfrac 1 {x_i + y_j}$.
+Subtract [[Definition:Column of Matrix|column]] $1$ from each of [[Definition:Column of Matrix|column]]s $2$ to $n$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a_{ij}
+      | o = \gets
+      | r = \frac 1 {x_i + y_j} - \frac 1 {x_i + y_1}
+      | c = 
+}}
+{{eqn | r = \frac {\left({x_i + y_1}\right) - \left({x_i + y_j}\right)} {\left({x_i + y_j}\right) \left({x_i + y_1}\right)}
+      | c = 
+}}
+{{eqn | r = \left({\frac {y_1 - y_j}{x_i + y_1} }\right) \left({\frac 1 {x_i + y_j} }\right)
+      | c = 
+}}
+{{end-eqn}}
+From [[Multiple of Row Added to Row of Determinant]] this will have no effect on the value of the [[Definition:Determinant of Matrix|determinant]].
+Now:
+:$1$: extract the factor $\dfrac 1 {x_i + y_1}$ from each [[Definition:Row of Matrix|row]] $1 \le i \le n$
+:$2$: extract the factor $y_1 - y_j$ from each [[Definition:Column of Matrix|column]] $2 \le j \le n$.
+Thus from [[Determinant with Row Multiplied by Constant]] we have the following:
+:$\displaystyle D_n = \left({\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j \mathop = 2}^n y_1 - y_j}\right) \begin{vmatrix}
+1 & \dfrac 1 {x_1 + y_2} & \dfrac 1 {x_1 + y_3} & \cdots & \dfrac 1 {x_1 + y_n} \\
+1 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
+1 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+1 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
+\end{vmatrix}$
+Now subtract [[Definition:Row of Matrix|row]] $1$ from each of [[Definition:Row of Matrix|row]]s $2$ to $n$.
+[[Definition:Column of Matrix|Column]] $1$ will go to $0$ for all but the first [[Definition:Row of Matrix|row]].
+[[Definition:Column of Matrix|Column]]s $2$ to $n$ will become:
+{{begin-eqn}}
+{{eqn | l = a_{ij}
+      | o = \gets
+      | r = \frac 1 {x_i + y_j} - \frac 1 {x_1 + y_j}
+      | c = 
+}}
+{{eqn | r = \frac {\left({x_1 + y_j}\right) - \left({x_i + y_j}\right)} {\left({x_i + y_j}\right) \left({x_1 + y_j}\right)}
+      | c = 
+}}
+{{eqn | r = \left({\frac {x_1 - x_i} {x_1 + y_j} }\right) \left({\frac 1 {x_i + y_j} }\right)
+      | c = 
+}}
+{{end-eqn}}
+From [[Multiple of Row Added to Row of Determinant]] this will have no effect on the value of the [[Definition:Determinant of Matrix|determinant]].
+Now:
+:$1$: extract the factor $x_1 - x_i$ from each [[Definition:Row of Matrix|row]] $2 \le i \le n$
+:$2$: extract the factor $\dfrac 1 {x_1 + y_j}$ from each [[Definition:Column of Matrix|column]] $2 \le j \le n$.
+Thus from [[Determinant with Row Multiplied by Constant]] we have the following:
+:$\displaystyle D_n = \left({\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j \mathop = 1}^n \frac 1 {x_1 + y_j}}\right) \left({\prod_{i \mathop = 2}^n x_1 - x_i}\right) \left({\prod_{j \mathop = 2}^n y_1 - y_j}\right) \begin{vmatrix}
+1 & 1 & 1 & \cdots & 1 \\
+0 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
+0 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+0 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
+\end{vmatrix}$
+From [[Determinant with Unit Element in Otherwise Zero Row]], and tidying up the products, we get:
+:$D_n = \frac {\displaystyle \prod_{i \mathop = 2}^n \left({x_i - x_1}\right) \left({y_i - y_1}\right)} {\displaystyle \prod_{1 \mathop \le i, j \mathop \le n} \left({x_i + y_1}\right) \left({x_1 + y_j}\right)}
+\begin{vmatrix}
+\dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
+\dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
+\end{vmatrix}$
+Repeat the process for the remaining [[Definition:Row of Matrix|row]]s and [[Definition:Column of Matrix|column]]s $2$ to $n$.
+The result follows.
+{{qed}}
+A similar process obtains the result for the $a_{ij} = \dfrac 1 {x_i - y_j}$ form.
+\end{proof}<|endoftext|>
+\section{Integers under Addition form Abelian Group}
+Tags: Integer Addition, Additive Group of Integers, Examples of Abelian Groups
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Addition|addition]] $\struct {\Z, +}$ forms an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Integer/Formal Definition|integers]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, +}$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] copy of the [[Definition:Inverse Completion|inverse completion]] of $\struct {\N, +}$.
+From [[Natural Numbers under Addition form Commutative Semigroup]], $\struct {\N, +}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+From [[Natural Number Addition is Cancellable]] all elements of $\struct {\N, +}$ are [[Definition:Cancellable Element|cancellable]].
+The result follows from [[Inverse Completion of Commutative Semigroup is Abelian Group]].
+Thus addition on $\Z$ is [[Definition:Well-Defined Operation|well-defined]], [[Definition:Closed Algebraic Structure|closed]], [[Definition:Associative|associative]] and [[Definition:Commutative Operation|commutative]] on $\Z$.
+{{qed|lemma}}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by:
+:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+=== Identity is Zero ===
+From [[Construction of Inverse Completion/Identity of Quotient Structure|Construction of Inverse Completion: Identity of Quotient Structure]], the [[Definition:Identity Element|identity]] of $\struct {\Z, +}$ is $\eqclass {c, c} {}$ for any $c \in \N$:
+{{begin-eqn}}
+{{eqn | ll= \forall a, b, c \in \N:
+      | o = 
+      | r = \eqclass {a, b} {} + \eqclass {c, c} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a, b} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {c, c} {} + \eqclass {a, b} {}
+      | c = 
+}}
+{{end-eqn|lemma}}
+$\eqclass {c, c} {}$ is the [[Definition:Equivalence Class|equivalence class]] of pairs of elements $\N \times \N$ whose [[Definition:Difference (Natural Numbers)|difference]] is [[Definition:Zero (Algebra)|zero]].
+Thus the [[Definition:Identity Element|identity]] of $\struct {\Z, +}$ is seen to be $0$.
+Note that a perfectly good representative of $\eqclass {c, c} {}$ is $\eqclass {0, 0} {}$.
+This usually keeps to a minimum the complexity of any arithmetic that is needed.
+{{qed|lemma}}
+=== Construction of Inverses ===
+From [[Construction of Inverse Completion/Invertible Elements in Quotient Structure|Construction of Inverse Completion: Invertible Elements in Quotient Structure]], we see that every element of $\struct {\Z, +}$ has an [[Definition:Inverse Element|inverse]].
+We can see that:
+{{begin-eqn}}
+{{eqn | ll= \forall a, b \in \N:
+      | l = \eqclass {a, b} {} + \eqclass {b, a} {}
+      | r = \eqclass {a + b, b + a} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a + b, a + b} {}
+      | c = [[Natural Number Addition is Commutative]]
+}}
+{{end-eqn}}
+The above construction is valid because $a$ and $b$ are both in $\N$ and hence [[Definition:Cancellable Element|cancellable]].
+From [[Construction of Inverse Completion/Identity of Quotient Structure|Construction of Inverse Completion: Identity of Quotient Structure]], $\eqclass {a + b, a + b} {}$ is a member of the equivalence class which is the [[Definition:Identity Element|identity]] of $\struct {\Z, +}$.
+Thus the [[Definition:Inverse Element|inverse]] of $\eqclass {a, b} {}$ is $\eqclass {b, a} {}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Piecewise Continuous Function with One-Sided Limits is Darboux Integrable}
+Tags: Piecewise Continuous Functions, Integral Calculus, Piecewise Continuous Function with One-Sided Limits is Darboux Integrable
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We are given that $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
+Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$ 
+:$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist.
+Note that $n$ is the number of [[Definition:Open Real Interval|intervals]] $\openint {x_{i - 1} } {x_i}$ defined from the [[Definition:Finite Subdivision|(finite) subdivision]] $\set {x_0, x_1, \ldots, x_n}$.
+We shall use [[Principle of Mathematical Induction|proof by induction]] on the [[Definition:Open Real Interval|intervals]].
+For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$.
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_{i - 1} } {x_i}$
+for an arbitrary $i \in \set {1, 2, \ldots, k}$. 
+[[Definition:Piecewise Continuous Function with One-Sided Limits|Piecewise continuity with one-sided limits]] of $f$ for the case $n = 1$ means that:
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$
+:$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist.
+By [[Integrability Theorem for Functions Continuous on Open Intervals]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_{i - 1} } {x_i}$.
+Thus $\map P 1$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\openint {x_0} {x_k}$.
+from which it is to be shown that:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\openint {x_0} {x_{k + 1} }$.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By definition of a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]], for every $i \in \set {1, 2, \ldots, k, k + 1}$:
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$
+:the [[Definition:One-Sided Limit of Real Function|one-sided limits]] $\displaystyle \lim_{x \mathop \to x_{i - 1}^+} \map f x$ and $\displaystyle \lim_{x \mathop \to x_i^-} \map f x$ exist.
+By the [[Piecewise Continuous Function with One-Sided Limits is Darboux Integrable#Induction Hypothesis|induction hypothesis]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$.
+From the [[Piecewise Continuous Function with One-Sided Limits is Darboux Integrable#Basis for the Induction|basis for the induction]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_k} {x_{k + 1} }$.
+We have that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$.
+Therefore, by [[Existence of Integral on Union of Adjacent Intervals]]:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$.
+We have that:
+:$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$
+Accordingly, $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_{k + 1} }$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We are given that $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on $\closedint a b$.
+From [[Piecewise Continuous Function with One-Sided Limits is Bounded]], $f$ is a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]].
+The result follows from [[Bounded Piecewise Continuous Function is Darboux Integrable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Schröder Rule/Proof 1}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $A$, $B$ and $C$ be [[Definition:Endorelation|relations]] on a [[Definition:Set|set]] $S$.
+Then the following are [[Definition:Logical Equivalence|equivalent]] statements:
+:$(1): \quad A \circ B \subseteq C$
+:$(2): \quad A^{-1} \circ \overline{C} \subseteq \overline{B}$
+:$(3): \quad \overline{C} \circ B^{-1} \subseteq \overline{A}$
+where:
+: $\circ$ denotes [[Definition:Composition of Relations|relation composition]]
+: $A^{-1}$ denotes the [[Definition:Inverse Relation|inverse]] of $A$
+: $\overline{A}$ denotes the [[Definition:Complement of Relation|complement]] of $A$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ iff $(2)$ ===
+By the definition of [[Definition:Composition of Relations|relation composition]] and [[Definition:Subset|subset]] we have that statement $(1)$ may be written as:
+:$(1'):\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, y) \in B  \implies  (x, z) \in C }\right)$
+Using a different arrangement of variable names, statement $(2)$ can be written:
+:$(2'):\quad \forall x, y, z \in S: \left({ (z, y) \in A^{-1} \land (x, z) \in \overline C \implies (x, y) \in \overline B }\right)$
+By the definition of [[Definition:Inverse Relation|inverse relation]] and the [[Definition:Complement of Relation|complement]] of a relation we can rewrite this as:
+:$(2''):\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, z) \notin C)  \implies  (x, y) \notin B }\right)$
+We shall use the [[Method of Truth Tables/Proof of Interderivability|method of truth tables]].
+The two statements will be [[Definition:Logical Equivalence|equivalent]] [[Definition:Iff|iff]] the columns under the [[Definition:Main Connective|main connectives]], which is $\implies$ in each case, are identical.
+'''Statement 1:'''
+$\begin{array}{ccccc}
+((y, z) \in A & \land & (x, y) \in B) & \implies & (x, z) \in C \\
+\hline
+T & T & T & T & T \\
+T & T & T & F & F \\
+T & F & F & T & T \\
+T & F & F & T & F \\
+F & F & T & T & T \\
+F & F & T & T & F \\
+F & F & F & T & T \\
+F & F & F & T & F \\
+\end{array}$
+'''Statement 2:'''
+$\begin{array}{ccccc}
+((y, z) \in A & \land & (x, z) \notin C) & \implies & (x, y) \notin B \\
+\hline
+T & F & F & T & F \\
+T & T & T & F & F \\
+T & F & F & T & T \\
+T & T & T & T & T \\
+F & F & F & T & F \\
+F & F & T & T & F \\
+F & F & F & T & T \\
+F & F & T & T & T \\
+\end{array}$
+{{qed|lemma}}
+=== $(2)$ iff $(3)$ ===
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Condition for Well-Foundedness/Forward Implication}
+Tags: Condition for Well-Foundedness
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $\struct {S, \preceq}$ be [[Definition:Well-Founded|well-founded]].
+Then there is no [[Definition:Infinite Sequence|infinite sequence]] $\sequence {a_n}$ of [[Definition:Element|elements]] of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$.
+\end{theorem}
+
+\begin{proof}
+Suppose there exists an [[Definition:Sequence|infinite sequence]] $\sequence {a_n}$ in $S$ such that:
+:$\forall n \in \N: a_{n + 1} \prec a_n$
+We let $T = \set {a_0, a_1, a_2, \ldots}$.
+Clearly $T$ has no [[Definition:Minimal Element|minimal element]].
+Thus by definition $S$ is not [[Definition:Well-Founded|well-founded]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extendability Theorem for Function Continuous on Open Interval}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Continuous Real Function on Open Interval|continuous real function]] that is defined on an [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let $g$ be a [[Definition:Real Function|real function]] that satisfies:
+:$g$ is defined on $\closedint a b$
+:$g$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\closedint a b$
+:$g = f$ on $\openint a b$.
+Then $g$ exists {{iff}} $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Assume that $g$ exists.
+We need to prove that the limits $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.
+$g$ is [[Definition:Continuous Real Function at Point|continuous]] at the [[Definition:Endpoint of Real Interval|end points]] $a$ and $b$ of its [[Definition:Domain of Mapping|domain]] as $g$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\closedint a b$.
+$g$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$ and [[Definition:Left-Continuous at Point|left-continuous]] at $b$.
+This means that $\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$.
+In turn, this means that the expressions $\displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\displaystyle \lim_{x \mathop \to b^-} \map g x$ exist.
+The two $x$ parameters as being part of the two limiting processes in these expressions can be considered confined to $\openint a b$.
+Therefore, $g$ in these two expressions can be replaced by $f$ as $g = f$ on $\openint a b$.
+We conclude that $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.
+We need to prove that a [[Definition:Real Function|function]] $g$ with the properties listed in the theorem exists.
+Define $g = f$ on $\openint a b$.
+Define:
+:$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$
+:$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$
+$g$ is defined at $a$ and $b$ as ensured by the existence of the [[Definition:Limit from Right|limits on the right hand sides]] of these two equations.
+We know that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$.
+Therefore, $g$ too is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$ as $g = f$ there.
+It remains to show that $g$ is [[Definition:Continuous Real Function at Point|continuous]] at $a$ and $b$.
+$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$ per definition.
+$x$ can be considered as being an element of $\openint a b$ as the limiting process at $a$ requires $x$ to approach $a$ from above.
+$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$ per definition.
+$x$ can be considered as being an element of $\openint a b$ as the limiting process at $b$ requires $x$ to approach $b$ from below.
+Therefore, the two equations above can be written:
+:$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$
+:$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$
+as $g = f$ on $\openint a b$.
+We conclude that $g$ is [[Definition:Continuous Real Function at Point|continuous]] at $a$ and $b$ as these two equations are exactly the definitions of [[Definition:Continuous Real Function at Point|continuity]] for $g$ at respectively $a$ and $b$.
+{{qed}}
+[[Category:Continuous Functions]]
+gwexoblpllrjrg5fmsbn6p203oqt2xg
+\end{proof}<|endoftext|>
+\section{Definition:Trivial Ordering}
+Tags: Definitions: Order Theory
+
+\begin{theorem}
+The '''trivial ordering''' is an [[Definition:Ordering|ordering]] $\mathcal R$ defined on a [[Definition:Set|set]] $S$ by:
+:$\forall a, b \in S: a \mathrel{\mathcal R} b \iff a = b$
+That is, there is no [[Definition:Ordering|ordering]] defined on any two distinct [[Definition:Element|elements]] of the [[Definition:Set|set]] $S$.
+\end{theorem}<|endoftext|>
+\section{Integrability Theorem for Functions Continuous on Open Intervals}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ such that $a < b$.
+Let $f$ be [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$.
+Let the [[Definition:One-Sided Limit of Real Function|one-sided limits]] $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We start by showing that $f$ is [[Definition:Bounded Real-Valued Function|bounded]].
+A [[Definition:Real Function|real function]] $g$ defined on $\closedint a b$ exists by [[Extendability Theorem for Function Continuous on Open Interval]] such that:
+:$g$ equals $f$ on $\openint a b$
+:$g$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\closedint a b$
+Therefore:
+:$g$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$ by [[Continuous Function on Compact Subspace of Euclidean Space is Bounded]] and [[Closed Real Interval is Compact in Metric Space]]
+Accordingly, $g$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\openint a b$ as $\openint a b$ is a [[Definition:Subset|subset]] of $\closedint a b$.
+Therefore, $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\openint a b$ as $f$ equals $g$ on $\openint a b$.
+Also, $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the [[Definition:Set|set]] $\set {a, b}$ by the [[Definition:Max Function|maximum]] $\map {\max} {\size {\map f a}, \size {\map f b}}$.
+Therefore:
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$
+We have:
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint a b$
+Therefore:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$ by [[Bounded Function Continuous on Open Interval is Darboux Integrable]].
+{{qed}}
+[[Category:Integral Calculus]]
+jlrf0rmt5onda0cxxy2wyqtomufrz4y
+\end{proof}<|endoftext|>
+\section{Set of Rational Numbers whose Numerator Divisible by p is Closed under Multiplication}
+Tags: Prime Numbers, Rational Multiplication, Algebraic Closure
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $A_p$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] which, when expressed in [[Definition:Canonical Form of Rational Number|canonical form]] has a [[Definition:Numerator|numerator]] which is [[Definition:Divisor of Integer|divisible]] by $p$.
+Then $A_p$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Multiplication|rational multiplication]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in A_p$.
+Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
+:$n_1, n_2 \in \Z$
+:$d_1, d_2 \in \Z_{>0}$
+:$p n_1 \perp d_1, p n_2 \perp d_2$
+Then:
+{{begin-eqn}}
+{{eqn | l = a \times b
+      | r = \frac {p n_1} {d_1} \times \frac {p n_2} {d_2}
+      | c = 
+}}
+{{eqn | r = \frac {p n_1 p n_2} {d_1 d_2}
+      | c = {{Defof|Rational Multiplication}}
+}}
+{{eqn | r = \frac {p^2 \left({n_1 n_2}\right)} {d_1 d_2}
+      | c = 
+}}
+{{end-eqn}}
+From [[Euclid's Lemma for Prime Divisors]], if $p \mathop \backslash d_1 d_2$ then either $p \mathop \backslash d_1$ or $p \mathop \backslash d_2$.
+But neither of these is the case, so $p \nmid d_1 d_2$.
+Hence by [[Prime not Divisor implies Coprime]]:
+:$p \perp d_1 d_2$
+where $\perp$ denotes [[Definition:Coprime Integers|coprimeness]].
+So when $\dfrac {p^2 \left({n_1 n_2}\right)} {d_1 d_2}$ is expressed in [[Definition:Canonical Form of Rational Number|canonical form]], $p$ will still be a [[Definition:Divisor of Integer|divisor]] of the [[Definition:Numerator|numerator]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Less One divides Power Less One}
+Tags: Number Theory
+
+\begin{theorem}
+Let $q, n \in \Z_{>0}$.
+Then:
+:$\paren {q - 1} \divides \paren {q^n - 1}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Geometric Sequence]]:
+:$\displaystyle \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n - 1} q^k$
+That is:
+:$q^n - 1 = r \paren {q - 1}$
+where $r = 1 + q + q^2 + \cdots + q^{n - 1}$.
+As [[Integer Addition is Closed]] and [[Integer Multiplication is Closed]], it follows that $r \in \Z$.
+Hence the result by definition of [[Definition:Divisor of Integer|divisor of integer]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number Plus One divides Power Plus One iff Odd}
+Tags: Number Theory
+
+\begin{theorem}
+Let $q, n \in \Z_{>0}$.
+Then:
+:$\paren {q + 1} \divides \paren {q^n + 1}$
+{{iff}} $n$ is [[Definition:Odd Integer|odd]].
+In the above, $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+Let $n$ be [[Definition:Odd Integer|odd]].
+Then from [[Sum of Odd Positive Powers]]:
+:$\displaystyle q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$
+Let $n$ be [[Definition:Even Integer|even]].
+Consider the equation:
+:$q^n + 1 = 0$
+We have:
+:$\paren {-1}^n + 1 = 2$
+So $-1$ is not a root of $q^n + 1 = 0$.
+By [[Polynomial Factor Theorem]], $q + 1$ is not a [[Definition:Divisor of Integer|divisor]] of $q^n + 1$ when $n$ is [[Definition:Even Integer|even]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping Bounded on Union iff Bounded on Each Component/Real-Valued Function}
+Tags: Real Analysis, Boundedness
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real-Valued Function|real-valued function]].
+Then:
+: $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on the [[Definition:Set Union|union]] of a [[Definition:Finite|finite]] number of [[Definition:Set|sets]] within the [[Definition:domain|domain]] of $f$
+{{iff}}:
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on each of the sets.
+\end{theorem}
+
+\begin{proof}
+Let $S_i$, $i \in \left\{{1, \ldots, n}\right\}$, $n \in \N_{>0}$, denote sets within the [[Definition:domain|domain]] of $f$.
+Define $S = \displaystyle \bigcup_{i \mathop = 1}^{n} S_i$.
+Suppose first that $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on each of the [[Definition:Set|sets]] $S_i$.
+We need to prove that $f$ is bounded on $S$.
+Let $b_i$ be a [[Definition:Bound of Real-Valued Function|bound]] for $f$ on $S_i$, $i \in \left\{{1, \ldots, n}\right\}$.
+Let $b = \max \left({b_1, \ldots, b_n}\right)$ where $\max$ is the [[Definition:Max Operation/General Definition/Real Numbers|maximum operation]].
+We note that $b \ge b_i$ for every $i$ in $\left\{{1, \ldots, n}\right\}$.
+Now, by definition of [[Definition:Bound of Real-Valued Function|bound]], a number greater than or equal to a bound is also a bound.
+Therefore, since $b_i$ is a bound for $f$ on $S_i$ and $b \ge b_i$, $b$ is a bound for $f$ on $S_i$ for every $i$.
+Since $b$ is a [[Definition:Bound of Real-Valued Function|bound]] for $f$ on every [[Definition:Set|set]] $S_i$, $b$ is a bound for $f$ on the [[Definition:Set Union|union]] of the sets $S_i$.
+In other words, $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $S$.
+This finishes the "if" part of the proof.
+Now, suppose that $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $S$.
+We need to prove that $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on each of the [[Definition:Set|sets]] $S_i$, $i \in \left\{{1, \ldots, n}\right\}$.
+Since $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $S$, there is a [[Definition:Bound of Real-Valued Function|bound]] $K$ that satisfies $K \ge \vert f\left( {s} \right) \vert$ for every [[Definition:Element|element]] $s$ in $S$.
+Pick a [[Definition:Set|set]] $S_i$.
+Since $K \ge \vert f\left( {s} \right) \vert$ is true of every [[Definition:Element|element]] $s$ of $S$, this is also true of every element of $S_i$  as $S_i$ is a [[Definition:subset|subset]] of $S$.
+Consequently, $K$ is a [[Definition:Bound of Real-Valued Function|bound]] for $f$ on $S_i$.
+Since $i$ is arbitrary, we conclude that $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $S_i$ for every $i$ in $\left\{{1, \ldots, n}\right\}$.
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Boundedness]]
+tpyk1kucdrcysb9i50xore6nw3psmsa
+\end{proof}<|endoftext|>
+\section{Parity of Integer equals Parity of Positive Power}
+Tags: Integers
+
+\begin{theorem}
+Let $p \in \Z$ be an [[Definition:Integer|integer]].
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then $p$ is [[Definition:Even Integer|even]] {{iff}} $p^n$ is [[Definition:Even Integer|even]].
+That is, the [[Definition:Parity|parity]] of an [[Definition:Integer|integer]] equals the [[Definition:Parity|parity]] of all its [[Definition:Strictly Positive Integer|(strictly) positive]] [[Definition:Power (Algebra)|powers]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:For all $p \in \Z$, $p$ is [[Definition:Even Integer|even]] {{iff}} $p^n$ is [[Definition:Even Integer|even]].
+First it is worth confirming that $P \left({0}\right)$ does not hold:
+:$\forall p \in \Z: p^0 = 1$
+which is not [[Definition:Even Integer|even]] whatever the [[Definition:Parity|parity]] of $p$.
+$P \left({1}\right)$ is true, as this just says:
+: $p$ is [[Definition:Even Integer|even]] {{iff}} $p$ is [[Definition:Even Integer|even]]
+which is a [[Definition:Tautology|tautology]].
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the case:
+: $p$ is [[Definition:Even Integer|even]] {{iff}} $p^2$ is [[Definition:Even Integer|even]]
+which is demonstrated in [[Parity of Integer equals Parity of its Square]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+: $p$ is [[Definition:Even Integer|even]] {{iff}} $p^k$ is [[Definition:Even Integer|even]]
+Then we need to show:
+: $p$ is [[Definition:Even Integer|even]] {{iff}} $p^{k+1}$ is [[Definition:Even Integer|even]]
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $p$ be [[Definition:Even Integer|even]].
+By the [[Parity of Integer equals Parity of Positive Power#Induction Hypothesis|induction hypothesis]], $p^k$ is also [[Definition:Even Integer|even]].
+Then:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = 2 r
+      | c = for some $r \in \Z$
+}}
+{{eqn | l = p^k
+      | r = 2 s
+      | c = for some $s \in \Z$
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = p^{k+1}
+      | r = p \cdot p^k
+      | c = 
+}}
+{{eqn | r = \left({2 r}\right) \left({2 s}\right)
+      | c = 
+}}
+{{eqn | r = 2 \left({2 r s}\right)
+      | c = 
+}}
+{{end-eqn}}
+So $p^{k+1}$ is [[Definition:Even Integer|even]].
+Now suppose $p$ is not [[Definition:Even Integer|even]] (that is, [[Definition:Odd Integer|odd]]).
+By the [[Parity of Integer equals Parity of Positive Power#Induction Hypothesis|induction hypothesis]], $p^k$ is also not [[Definition:Even Integer|even]] (that is, [[Definition:Odd Integer|odd]]).
+Then:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = 2 r + 1
+      | c = for some $r \in \Z$
+}}
+{{eqn | l = p^k
+      | r = 2 s + 1
+      | c = for some $s \in \Z$
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = p^{k+1}
+      | r = p \cdot p^k
+      | c = 
+}}
+{{eqn | r = \left({2 r + 1}\right) \left({2 s + 1}\right)
+      | c = 
+}}
+{{eqn | r = 4 r s + 2 \left({r + s}\right) + 1
+      | c = 
+}}
+{{eqn | r = 2 \left({2 r s + r + s}\right) + 1
+      | c = 
+}}
+{{end-eqn}}
+So $p^{k+1}$ is not [[Definition:Even Integer|even]] (that is, [[Definition:Odd Integer|odd]]).
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:For all $n \in \Z_{>0}$, for all $p \in \Z$, $p$ is [[Definition:Even Integer|even]] {{iff}} $p^n$ is [[Definition:Even Integer|even]].
+{{qed}}
+[[Category:Integers]]
+4kflgjxhdq7h3hb59iyxn37ue7f0s11
+\end{proof}<|endoftext|>
+\section{Two divides Power Plus One iff Odd}
+Tags: Number Theory
+
+\begin{theorem}
+Let $q, n \in \Z_{>0}$.
+Then:
+:$2 \divides \paren {q^n + 1}$
+{{iff}} $q$ is [[Definition:Odd Integer|odd]].
+In the above, $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+By [[Parity of Integer equals Parity of Positive Power]], $q^n$ is [[Definition:Even Integer|even]] {{iff}} $q$ is [[Definition:Even Integer|even]].
+Thus it follows that $q^n + 1$ is [[Definition:Even Integer|even]] {{iff}} $q$ is [[Definition:Odd Integer|odd]].
+The result follows by definition of [[Definition:Even Integer|even integer]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primes of form Power plus One}
+Tags: Number Theory
+
+\begin{theorem}
+Let $q, n \in \Z_{>0}$ such that $q > 1$.
+Then $q^n + 1$ is [[Definition:Prime Number|prime]] only if:
+:$(1): \quad q$ is [[Definition:Even Integer|even]]
+and
+:$(2): \quad n$ is of the form $2^m$ for some [[Definition:Positive Integer|positive integer]] $m$.
+\end{theorem}
+
+\begin{proof}
+Note that if $q = 1$ then $q^n + 1 = 2$ which ''is'' [[Definition:Prime Number|prime]].
+Hence the condition on $q$ in the statement of the theorem.
+So [[Definition:By Hypothesis|by hypothesis]] $q > 1$.
+Let $q$ be [[Definition:Odd Integer|odd]].
+Then by [[Two divides Power Plus One iff Odd]], $q^n + 1$ is not [[Definition:Prime Number|prime]].
+Let $q$ be [[Definition:Even Integer|even]].
+Let $n$ be expressed in the form $r 2^m$ where $r$ is [[Definition:Odd Integer|odd]].
+Then $q^n + 1$ can be expressed in the form:
+: $q^{r 2^m} + 1 = \left({q^{2^m}}\right)^r + 1$
+By [[Number Plus One divides Power Plus One iff Odd]], $q^{2^m} + 1$ is a [[Definition:Divisor of Integer|divisor]] of $\left({q^{2^m}}\right)^r + 1$.
+So for all $r > 1$ it follows that $\left({q^{2^m}}\right)^r + 1$ is [[Definition:Composite Number|composite]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fermat Number whose Index is Sum of Integers}
+Tags: Fermat Numbers
+
+\begin{theorem}
+Let $F_n = 2^{\left({2^n}\right)} + 1$ be the $n$th [[Definition:Fermat Number|Fermat number]].
+Let $k \in \Z_{>0}$.
+Then:
+:$F_{n + k} - 1 = \left({F_n - 1}\right)^{2^k}$
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Fermat Number|Fermat number]]
+{{begin-eqn}}
+{{eqn | l = F_{n + k} - 1
+      | r = 2^{2^{n + k} }
+      | c = {{Defof|Fermat Number}}
+}}
+{{eqn | r = 2^{2^n 2^k}
+      | c = 
+}}
+{{eqn | r = \left({2^{2^n} }\right)^{2^k}
+      | c = 
+}}
+{{eqn | r = \left({F_n - 1}\right)^{2^k}
+      | c = {{Defof|Fermat Number}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Goldbach's Theorem}
+Tags: Fermat Numbers, Coprime Integers, Goldbach's Theorem
+
+\begin{theorem}
+Let $F_m$ and $F_n$ be [[Definition:Fermat Number|Fermat numbers]] such that $m \ne n$.
+Then $F_m$ and $F_n$ are [[Definition:Coprime Integers|coprime]].
+\end{theorem}<|endoftext|>
+\section{Number of Boolean Interpretations for Finite Set of Variables}
+Tags: Boolean Interpretations
+
+\begin{theorem}
+Let $\mathcal P_0$ be the [[Definition:Vocabulary of Propositional Logic|vocabulary]] of [[Definition:Language of Propositional Logic|language of propositional logic]].
+Let $S \subseteq \mathcal P_0$ be a [[Definition:Finite Set|finite set]] of $n$ [[Definition:Letter|letters]] from $\mathcal P_0$.
+Then there are $2^n$ different [[Definition:Partial Boolean Interpretation|partial boolean interpretations]] for $S$.
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Partial Boolean Interpretation|partial boolean interpretation]] for $S$ is a [[Definition:Mapping|mapping]] from $S$ to the [[Definition:Set of Truth Values|set of truth values]] $\set {T, F}$.
+By [[Cardinality of Set of All Mappings]], the total number of [[Definition:Mapping|mappings]] from $S$ to $T$ is:
+:$\card {T^S} = \card T^{\card S}$
+The result follows directly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Addition form Monoid}
+Tags: Integer Addition, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Addition|addition]] $\struct {\Z, +}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Integers under Addition form Abelian Group]].
+By definition, an [[Definition:Abelian Group|abelian group]] is a [[Definition:Group|group]].
+Also by definition, a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers under Addition form Monoid}
+Tags: Rational Addition, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] under [[Definition:Rational Addition|addition]] $\struct {\Q, +}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Rational Numbers under Addition form Abelian Group]].
+By definition, an [[Definition:Abelian Group|abelian group]] is a [[Definition:Group|group]].
+Also by definition, a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Field}
+Tags: Examples of Fields, Real Numbers
+
+\begin{theorem}
+The [[Definition:Real Number|set of real numbers]] $\R$ forms a [[Definition:Field (Abstract Algebra)|field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Numbers under Addition form Abelian Group]], we have that $\struct {\R, +}$ forms an [[Definition:Abelian Group|abelian group]].
+From [[Non-Zero Real Numbers under Multiplication form Abelian Group]], we have that $\struct {\R_{\ne 0}, \times}$ forms an [[Definition:Abelian Group|abelian group]].
+Next we have that [[Real Multiplication Distributes over Addition]].
+Thus all the criteria are fulfilled, and $\struct {\R, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Field}
+Tags: Examples of Fields, Rational Numbers
+
+\begin{theorem}
+Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Q, +, \times}$, where:
+:$\Q$ is the set of all [[Definition:Rational Number|rational numbers]]
+:$+$ is the operation of [[Definition:Rational Addition|rational addition]]
+:$\times$ is the operation of [[Definition:Rational Multiplication|rational multiplication]].
+Then $\struct {\Q, +, \times}$ forms a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+This is demonstrated in the [[Definition:Rational Number/Formal Definition|formal definition of rational numbers]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers under Addition form Monoid}
+Tags: Real Addition, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] under [[Definition:Real Addition|addition]] $\struct {\R, +}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Monoid Axioms|monoid axioms]] in turn:
+=== $\text S 0$: Closure ===
+[[Real Addition is Closed]].
+{{qed|lemma}}
+=== $\text S 1$: Associativity ===
+[[Real Addition is Associative]].
+{{qed|lemma}}
+=== $\text S 2$: Identity ===
+From [[Real Addition Identity is Zero]], we have that the [[Definition:Identity Element|identity element]] of $\struct {\R, +}$ is the [[Definition:Real Number|real number]] $0$.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers Modulo m under Addition form Abelian Group}
+Tags: Additive Groups of Integers Modulo m, Modulo Addition, Examples of Abelian Groups
+
+\begin{theorem}
+Let $\Z_m$ is the set of [[Definition:Integers Modulo m|integers modulo $m$]]
+Let $+_m$ be the [[Definition:Binary Operation|operation]] of [[Definition:Modulo Addition|addition modulo $m$]].
+Then the [[Definition:Algebraic Structure|structure]] $\struct {\Z_m, +_m}$ is an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers Modulo m under Addition form Cyclic Group]], $\struct {\Z_m, +_m}$ is a [[Definition:Cyclic Group|cyclic group]].
+The result follows from [[Cyclic Group is Abelian]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers under Addition form Monoid}
+Tags: Complex Addition, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] under [[Definition:Complex Addition|addition]] $\left({\C, +}\right)$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Monoid Axioms|monoid axioms]] in turn:
+=== $\text S 0$: Closure ===
+[[Complex Addition is Closed]].
+{{qed|lemma}}
+=== $\text S 1$: Associativity ===
+[[Complex Addition is Associative]].
+{{qed|lemma}}
+=== $\text S 2$: Identity ===
+From [[Complex Addition Identity is Zero]], we have that the [[Definition:Identity Element|identity element]] of $\struct {\C, +}$ is the [[Definition:Complex Number|complex number]] $0 + 0 i$:
+:$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$
+and similarly for $\paren {0 + 0 i} + \paren {x + i y}$.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers under Multiplication form Monoid}
+Tags: Rational Multiplication, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] under [[Definition:Rational Multiplication|multiplication]] $\struct {\Q, \times}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Monoid Axioms|monoid axioms]] in turn:
+=== $\text S 0$: Closure ===
+[[Rational Multiplication is Closed]].
+{{qed|lemma}}
+=== $\text S 1$: Associativity ===
+[[Rational Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text S 2$: Identity ===
+[[Rational Multiplication Identity is One|Rational Multiplication Identity is $1$]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers under Multiplication form Monoid}
+Tags: Real Multiplication, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] under [[Definition:Real Multiplication|multiplication]] $\struct {\R, \times}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Monoid Axioms|monoid axioms]] in turn:
+=== $\text S 0$: Closure ===
+[[Real Multiplication is Closed]].
+{{qed|lemma}}
+=== $\text S 1$: Associativity ===
+[[Real Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text S 2$: Identity ===
+[[Real Multiplication Identity is One|Real Multiplication Identity is $1$]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers under Multiplication form Monoid}
+Tags: Complex Multiplication, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] under [[Definition:Complex Multiplication|multiplication]] $\struct {\C, \times}$ forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Monoid Axioms|monoid axioms]] in turn:
+=== $\text S 0$: Closure ===
+[[Complex Multiplication is Closed]].
+{{qed|lemma}}
+=== $\text S 1$: Associativity ===
+[[Complex Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text S 2$: Identity ===
+[[Complex Multiplication Identity is One|Complex Multiplication Identity is $1$]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Natural Numbers under Multiplication form Commutative Monoid}
+Tags: Natural Numbers, Examples of Monoids
+
+\begin{theorem}
+Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], i.e. $\N_{>0} = \N \setminus \set 0$.
+The [[Definition:Algebraic Structure|structure]] $\struct{\N_{>0}, \times}$ forms a [[Definition:Commutative Monoid|commutative monoid]].
+\end{theorem}
+
+\begin{proof}
+From [[Non-Zero Natural Numbers under Multiplication form Commutative Semigroup]], $\struct {\N_{>0}, \times}$ forms a [[Definition:Commutative Semigroup|commutative semigroup]].
+From [[Identity Element of Natural Number Multiplication is One]], $\struct {\N_{>0}, \times}$ has an [[Definition:Identity Element|identity element]] which is $1$.
+Hence the result, by definition of [[Definition:Commutative Monoid|commutative monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Natural Numbers under Addition do not form Monoid}
+Tags: Natural Numbers, Monoids
+
+\begin{theorem}
+Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], i.e. $\N_{>0} = \N \setminus \left\{{0}\right\}$.
+The [[Definition:Algebraic Structure|structure]] $\left({\N_{>0}, +}\right)$ does ''not'' form a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+From [[Natural Numbers under Addition form Commutative Monoid]], $\left({\N, +}\right)$ forms a [[Definition:Commutative Monoid|commutative monoid]].
+From [[Natural Numbers Bounded Below under Addition form Commutative Semigroup]], $\left({\N_{>0}, +}\right)$ forms a [[Definition:Commutative Semigroup|commutative semigroup]].
+From [[Identity Element of Natural Number Addition is Zero]], $0$ is the [[Definition:Identity Element|identity]] of $\left({\N, +}\right)$ 
+From [[Natural Number Addition is Cancellable]], all elements of $\left({\N, +}\right)$ are [[Definition:Cancellable Element|cancellable]].
+Suppose $\left({\N_{>0}, +}\right)$ is a [[Definition:Monoid|monoid]].
+From [[Identity of Cancellable Monoid is Identity of Submonoid]], if $\left({\N_{>0}, +}\right)$ has an [[Definition:Identity Element|identity]] then it is the same as the [[Definition:Identity Element|identity]] of $\left({\N, +}\right)$.
+But as $0 \notin \N_{>0}$, it follows that $\left({\N_{>0}, +}\right)$ has no [[Definition:Identity Element|identity]].
+Hence $\left({\N_{>0}, +}\right)$ is not a [[Definition:Monoid|monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers Bounded Below under Addition form Commutative Semigroup}
+Tags: Examples of Semigroups, Natural Numbers
+
+\begin{theorem}
+Let $m \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
+Let $M \subseteq \N$ be defined as:
+:$M := \set {x \in \N: x \ge m}$
+That is, $M$ is the set of all [[Definition:Natural Numbers|natural numbers]] greater than or equal to $m$.
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {M, +}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Natural Number Addition is Associative]]
+:[[Natural Number Addition is Commutative]]
+From [[Restriction of Associative Operation is Associative]], $+$ is [[Definition:Associative|associative]] on $\struct {M, +}$.
+From [[Restriction of Commutative Operation is Commutative]], $+$ is [[Definition:Commutative Operation|commutative]] on $\struct {M, +}$.
+It remains to be shown that $+$ is [[Definition:Closed Operation|closed]] on $\struct {M, +}$.
+Let $a, b \in M$.
+Then $\exists r, s \in \N: a = m + r, b = m + s$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a + b
+      | r = \paren {m + r} + \paren {m + s}
+      | c = 
+}}
+{{eqn | r = m + \paren {m + r + s}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = m
+      | c = 
+}}
+{{end-eqn}}
+So $a + b \in M$ and so $+$ is [[Definition:Closed Operation|closed]] on $\struct {M, +}$.
+{{qed}}
+[[Category:Examples of Semigroups]]
+[[Category:Natural Numbers]]
+miqbplbsv6ao8s0vmpg3sajbedkaty7
+\end{proof}<|endoftext|>
+\section{Positive Real Numbers under Max Operation form Monoid}
+Tags: Real Numbers, Max and Min Operations, Examples of Monoids
+
+\begin{theorem}
+Let $\R_{\ge 0}$ be the set of [[Definition:Positive Real Number|positive (that is, non-negative) real numbers]].
+Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the [[Definition:Max Operation|max operation]] on $\R_{\ge 0}$.
+Then $\left({\R_{\ge 0}, \max}\right)$ is a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $0$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Numbers are Totally Ordered]], $\R$ is a [[Definition:Totally Ordered Set|totally ordered set]].
+From [[Max Operation on Toset is Semigroup]], $\left({\R_{\ge 0}, \max}\right)$ is a [[Definition:Semigroup|semigroup]].
+By definition of $\R_{\ge 0}$:
+:$\forall x \in \R_{\ge 0}: 0 \le x$
+Thus by definition of the [[Definition:Max Operation|max operation]]:
+:$\forall x \in \R_{\ge 0}: \max \left({0, x}\right) = x = \max \left({x, 0}\right)$
+So $0$ is the [[Definition:Identity Element|identity]] of $\left({\R_{\ge 0}, \max}\right)$ by definition.
+The result follows by definition of [[Definition:Monoid|monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Free Commutative Monoid on One Element is Isomorphic to Positive Integers under Addition}
+Tags: Natural Numbers, Free Monoids
+
+\begin{theorem}
+Let $X = \left\{{x}\right\}$ be a [[Definition:Singleton|singleton]].
+Let $M$ be the [[Definition:Free Commutative Monoid|free commutative monoid]] on $X$.
+Then $M$ is [[Definition:Monoid Isomorphism|isomorphic]] to the [[Definition:Additive Monoid of Natural Numbers|additive monoid of natural numbers]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Free Commutative Monoid|free commutative monoid]] $M$ on $\left\{{x}\right\}$ is:
+:$M = \left\{{e, x, x^2, x^3, \ldots}\right\}$
+where $e$ denotes the [[Definition:Null Sequence|null sequence]] of elements of $X$.
+Let $\phi$ denote the [[Definition:Mapping|mapping]] from $M$ to $\N$ as:
+:$\forall a \in M: \phi \left({a}\right) = \begin{cases}
+0 & : a = e \\
+n & : a = x^n \end{cases}$
+By definition of $\phi$:
+:$\phi$ is [[Definition:Injection|injective]]: $\phi \left({a}\right) = \phi \left({b}\right) \implies a = b$
+:$\phi$ is [[Definition:Surjection|surjective]]: $\forall a \in \N: \exists b \in M: \phi \left({b}\right) = a$
+:$\phi$ is a [[Definition:Monoid Homomorphism|monoid homomorphism]]: $\phi \left({a b}\right) = \phi \left({a + b}\right) = \phi \left({a}\right) + \phi \left({b}\right)$
+Hence the result, by definition of [[Definition:Monoid Isomorphism|isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Monoid with Right Cancellable Operation is Group}
+Tags: Monoids, Finite Groups
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Finite Monoid|finite monoid]].
+Let $\circ$ be a [[Definition:Right Cancellable Operation|right cancellable operation]].
+Then $\struct {S, \circ}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:Group Axioms|Group axioms $\text G 0$, $\text G 1$ and $\text G 2$]] are satisfied by dint of $\struct {S, \circ}$ being a [[Definition:Monoid|monoid]].
+Recall the definition of [[Definition:Right Cancellable Operation|right cancellable operation]]:
+:$\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$
+Let $\rho_c: S \to S$ be the [[Definition:Right Regular Representation|right regular representation]] of $\struct {S, \circ}$ with respect to $c$.
+By [[Right Cancellable iff Right Regular Representation Injective]], $\rho_c$ is an [[Definition:Injection|injection]].
+By [[Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection]], $\rho_c$ is a [[Definition:Bijection|bijection]].
+Thus $a \circ b = e$ has a [[Definition:Unique|unique]] solution for all $a \in S$.
+That is, [[Definition:Group Axioms|group axiom $\text G 3$]] holds on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Monoid with Left Cancellable Operation is Group}
+Tags: Monoids, Finite Groups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Finite Monoid|finite monoid]].
+Let $\circ$ be a [[Definition:Left Cancellable Operation|left cancellable operation]].
+Then $\left({S, \circ}\right)$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:Group Axioms|Group axioms $G0$, $G1$ and $G2$]] are satisfied by dint of $\left({S, \circ}\right)$ being a [[Definition:Monoid|monoid]].
+Recall the definition of [[Definition:Left Cancellable Operation|left cancellable operation]]:
+:$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$
+Let $\lambda_c: S \to S$ be the [[Definition:Left Regular Representation|left regular representation]] of $\left({S, \circ}\right)$ with respect to $c$.
+By [[Left Cancellable iff Left Regular Representation Injective]], $\lambda_c$ is an [[Definition:Injection|injection]].
+By [[Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection]], $\lambda_c$ is a [[Definition:Bijection|bijection]].
+Thus $a \circ b = e$ has a [[Definition:Unique|unique]] solution for all $a \in S$.
+That is, [[Definition:Group Axioms|group axiom $G3$]] holds on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Field is Abelian Group}
+Tags: Field Theory, Abelian Groups, Multiplicative Group of Field is Abelian Group
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $F^* := F \setminus \set 0$ be the set $F$ less its [[Definition:Field Zero|zero]].
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {F^*, \times}$ is an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Field Axioms|field axioms]]:
+{{begin-axiom}}
+{{axiom | n = \text M 0
+        | lc= [[Definition:Closed Algebraic Structure|Closure]] under [[Definition:Field Product|product]]
+        | q = \forall x, y \in F
+        | m = x \circ y \in F
+}}
+{{axiom | n = \text M 1
+        | lc= [[Definition:Associative|Associativity]] of [[Definition:Field Product|product]]
+        | q = \forall x, y, z \in F
+        | m = \paren {x \circ y} \circ z = x \circ \paren {y \circ z}
+}}
+{{axiom | n = \text M 2
+        | lc= [[Definition:Commutative Operation|Commutativity]] of [[Definition:Field Product|product]]
+        | q = \forall x, y \in F
+        | m = x \circ y = y \circ x
+}}
+{{axiom | n = \text M 3
+        | lc= [[Definition:Identity Element|Identity element]] for [[Definition:Field Product|product]]
+        | q = \exists 1_F \in F, 1_F \ne 0_F: \forall x \in F
+        | m = x \circ 1_F = x = 1_F \circ x
+        | rc= $1_F$ is called the [[Definition:Unity of Field|unity]]
+}}
+{{axiom | n = \text M 4
+        | lc= [[Definition:Inverse Element|Inverse elements]] for [[Definition:Field Product|product]]
+        | q = \forall x \in F^*: \exists x^{-1} \in F^*
+        | m = x \circ x^{-1} = 1_F = x^{-1} \circ x
+}}
+{{end-axiom}}
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Commutative Division Ring|commutative division ring]].
+The result follows from [[Non-Zero Elements of Division Ring form Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One and Minus One form Subgroup of Multiplicative Group of Rational Numbers}
+Tags: Rational Numbers, Subgroups
+
+\begin{theorem}
+Let $\left({\Q_{\ne 0}, \times}\right)$ be the [[Definition:Multiplicative Group of Rational Numbers|multiplicative group of rational numbers]].
+Let $S \subseteq \Q$ where $S = \left\{{1, -1}\right\}$.
+Then $\left({S, \times}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({\Q_{\ne 0}, \times}\right)$.
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], $S$ is not [[Definition:Empty Set|empty]].
+As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$.
+Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$.
+Thus:
+:$\forall x \in S: x \times y^{-1} \in S$
+The result follows from the [[One-Step Subgroup Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetric Group is not Abelian}
+Tags: Symmetric Groups, Symmetric Group is not Abelian
+
+\begin{theorem}
+Let $S_n$ be the [[Definition:Symmetric Group|symmetric group]] of order $n$ where $n \ge 3$.
+Then $S_n$ is not [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c \in S$.
+Let $\alpha$ be the [[Definition:Transposition|transposition]] on $S$ which exchanges $a$ and $b$.
+Let $\beta$ be the [[Definition:Transposition|transposition]] on $S$ which exchanges $b$ and $c$.
+Then:
+:$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$
+while:
+:$\beta \circ \alpha$ maps $\tuple {a, b, c}$ to $\tuple {b, c, a}$
+Thus $\alpha, \beta \in S_n$ such that $\alpha$ does not [[Definition:Commute|commute]] with $\beta$.
+Hence the result by definition of [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\alpha \in S_n$ such that $\alpha$ is not the [[Definition:Identity Mapping|identity mapping]].
+From [[Center of Symmetric Group is Trivial]], $\alpha$ is not in the [[Definition:Center of Group|center]] $Z \paren {S_n}$ of $S_n$.
+Thus $S_n \ne Z \paren {S_n}$.
+The result follows by [[Group equals Center iff Abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Möbius Function is Multiplicative/Corollary}
+Tags: Multiplicative Functions, Möbius Function
+
+\begin{theorem}
+Let $\gcd \set {m, n} > 1$.
+Then:
+:$\map \mu {m n} = 0$
+\end{theorem}
+
+\begin{proof}
+Let $\gcd \set {m, n} = k$ where $k > 1$.
+Then $m = k r$ and $n = k s$ for some $r, s \in \Z$.
+Thus $m n = k^2 r s$.
+From [[Integer is Expressible as Product of Primes]] there exists $p \in \Z$ such that $p$ is [[Definition:Prime Number|prime]] and $p \divides k$.
+That is:
+:$\exists t \in \Z: k = p t$
+and so:
+:$m n = p^2 t^2 r s$
+That is:
+:$p^2 \divides m $
+Hence the result by definition of the [[Definition:Möbius Function|Möbius function]].
+{{qed}}
+[[Category:Multiplicative Functions]]
+[[Category:Möbius Function]]
+oxry1u2cq5rmz4yamh7ls14ie4hb64z
+\end{proof}<|endoftext|>
+\section{Affine Group of One Dimension is Group}
+Tags: Affine Groups
+
+\begin{theorem}
+Let $\map {\operatorname {Af}_1} \R$ be the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]].
+Then $\map {\operatorname {Af}_1} \R$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+Let $f_{ab}, f_{cd} \in \map {\operatorname {Af}_1} \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_{ab} \circ f_{cd} } x
+      | r = a \paren {c x + d} + b
+      | c = 
+}}
+{{eqn | r = a c x + a d + b
+      | c = 
+}}
+{{end-eqn}}
+By the [[Definition:Field Axioms|field axioms]], $a c \in \R_{\ne 0}$ and $a d + b \in \R$.
+Thus $f_{ab} \circ f_{cd} \in \map {\operatorname {Af}_1} \R$ and so $\map {\operatorname {Af}_1} \R$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+From [[Composition of Mappings is Associative]], it follows directly that $\circ$ is [[Definition:Associative|associative]] on $\map {\operatorname {Af}_1} \R$.
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+By [[Identity of Affine Group of One Dimension]], $\map {\operatorname {Af}_1} \R$ has $f_{1, 0}$ as an [[Definition:Identity Element|identity element]].
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+By [[Inverse in Affine Group of One Dimension]], every element $f_{a b}$ of $\map {\operatorname {Af}_1} \R$ has an [[Definition:Inverse Element|inverse]] $f_{c d}$ where $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
+{{qed|lemma}}
+All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\map {\operatorname {Af}_1} \R$ is a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity of Affine Group of One Dimension}
+Tags: Affine Groups
+
+\begin{theorem}
+The [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]] $\operatorname{Af}_1 \left({\R}\right)$ has $f_{1, 0}$ as an [[Definition:Identity Element|identity element]].
+\end{theorem}
+
+\begin{proof}
+Let $f_{ab} \in \operatorname{Af}_1 \left({\R}\right)$.
+Then:
+{{begin-eqn}}
+{{eqn | l = f_{ab} \circ f_{1, 0} \left({x}\right)
+      | r = a \left({1 x + 0}\right) + b
+      | c = 
+}}
+{{eqn | r = a x + b
+      | c = 
+}}
+{{eqn | r = f_{ab}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_{1, 0} \circ f_{a b} \left({x}\right)
+      | r = 1 \left({a x + b}\right) + 0
+      | c = 
+}}
+{{eqn | r = a x + b
+      | c = 
+}}
+{{eqn | r = f_{ab}
+      | c = 
+}}
+{{end-eqn}}
+Thus $f_{1, 0}$ is the [[Definition:Identity Element|identity element]] of $\operatorname{Af}_1 \left({\R}\right)$.
+{{qed}}
+[[Category:Affine Groups]]
+qg1cgk88mnkxd2maqjzdexhsrxhcwh9
+\end{proof}<|endoftext|>
+\section{Inverse in Affine Group of One Dimension}
+Tags: Affine Groups
+
+\begin{theorem}
+Let $\map {\operatorname {Af}_1} \R$ denote the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]].
+Let $f_{a b} \in \map {\operatorname {Af}_1} \R$.
+Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
+Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the [[Definition:Inverse Element|inverse]] of $f_{a b}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = a x + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y - b
+      | r = a x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y - b} a
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac y a + \frac {- b} a
+      | r = x
+      | c = 
+}}
+{{end-eqn}}
+As $a \in \R_{\ne 0}$ by definition of $\map {\operatorname {Af}_1} \R$ it follows that $\dfrac 1 a \in \R_{\ne 0}$ and $\dfrac {-b} a \in \R$.
+So let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_{a b} \circ f_{c d} } x
+      | r = a \paren {\frac 1 a x + \frac {-b} a} + b
+      | c = 
+}}
+{{eqn | r = \frac a a x + \frac {-a b} a + b
+      | c = 
+}}
+{{eqn | r = 1 x + 0
+      | c = 
+}}
+{{eqn | r = \map {f_{1, 0} } x
+      | c = 
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \map {f_{c d} \circ f_{a b} } x
+      | r = \frac 1 a \paren {a x + b} + \frac {-b} a
+      | c = 
+}}
+{{eqn | r = \frac a a x + \frac b a + \frac {-b} a
+      | c = 
+}}
+{{eqn | r = 1 x + 0
+      | c = 
+}}
+{{eqn | r = \map {f_{1, 0} } x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Commutation with Group Elements implies Commuation with Product with Inverse}
+Tags: Group Theory, Commutativity
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $a, b, c \in G$ such that $a$ [[Definition:Commute|commutes]] with both $b$ and $c$.
+Then $a$ [[Definition:Commute|commutes]] with $b c^{-1}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a b c^{-1}
+      | r = b a c^{-1}
+      | c = as $a$ [[Definition:Commute|commutes]] with $b$
+}}
+{{eqn | r = b c^{-1} a
+      | c = [[Commutation with Inverse in Monoid]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Möbius Strip has Euler Characteristic Zero}
+Tags: 
+
+\begin{theorem}
+Let $M$ be a [[Definition:Möbius Strip|Möbius Strip]].
+Then:
+:$\map \chi M = 0$
+where $\map \chi M$ denotes the [[Definition:Euler Characteristic|Euler characteristic]] of the [[Definition:Graph (Graph Theory)|graph]] $M$.
+\end{theorem}
+
+\begin{proof}
+Let the number of [[Definition:Vertex of Graph|vertices]], [[Definition:Edge of Graph|edges]] and [[Definition:Face of Graph|faces]] of $M$ be $V$, $E$ and $F$ respectively.
+From [[Möbius Strip has no Vertices]]:
+:$V = 0$
+From [[Möbius Strip has 1 Edge]]:
+:$E = 1$
+From [[Möbius Strip has 1 Face]]:
+:$F = 1$
+By definition of the [[Definition:Euler Characteristic|Euler characteristic]]:
+{{begin-eqn}}
+{{eqn | l = \map \chi M
+      | r = V - E + F
+      | c = Definition of [[Definition:Euler Characteristic|Euler Characteristic]]
+}}
+{{eqn | r = 0 - 1 + 1
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Centralizer of Subset is Intersection of Centralizers of Elements}
+Tags: Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S \subseteq G$.
+Let $\map {C_G} S$ be the [[Definition:Centralizer of Group Subset|centralizer of $S$ in $G$]].
+Then:
+:$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$
+where $\map {C_G} z$ is the [[Definition:Centralizer of Group Element|centralizer of $x$ in $G$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {C_G} S
+      | r = \set {g \in G: \forall x \in S: g \circ x = x \circ g}
+      | c = {{Defof|Centralizer of Group Subset}}
+}}
+{{eqn | r = \set {g \in G: \forall x \in S: g \in \map {C_G} x}
+      | c = {{Defof|Centralizer of Group Element}}
+}}
+{{eqn | r = \bigcap_{x \mathop \in S} \map {C_G} x
+      | c = {{Defof|Intersection of Family}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Permutation Representation defines Group Action}
+Tags: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $X$ be a [[Definition:Set|set]].
+Let $\map \Gamma X$ be the [[Definition:Symmetric Group|symmetric group]] of $X$.
+Let $\rho: G \to \map \Gamma X$ be a [[Definition:Permutation Representation|permutation representation]], that is, a [[Definition:Group Homomorphism|homomorphism]].
+The [[Definition:Mapping|mapping]] $\phi: G \times X \to X$ [[Definition:Group Action/Permutation Representation|associated to]] $\rho$, defined by:
+:$\map \phi {g, x} = \map {\map \rho g} x$
+is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+Let $g, h \in G$ and $x \in X$.
+We verify that $g * \paren {h * x} = \paren {g h} * x$:
+{{begin-eqn}}
+{{eqn | l = g * \paren {h * x}
+      | r = \map {\map \rho g} {\map {\map \rho h} x}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \map {\paren {\map \rho g \circ \map \rho h} } x
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\map \rho {g h} } x
+      | c = $\rho$ is a [[Definition:Group Homomorphism|homomorphism]]
+}}
+{{eqn | r = \paren {g h} * x
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+We verify that $e * x = x$.
+Let $I_X$ denote the [[Definition:Identity Mapping|identity mapping]] on $X$.
+{{begin-eqn}}
+{{eqn | l = e*x
+      | r = \map {\map \rho e} x
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \map {I_X} x
+      | c = [[Group Homomorphism Preserves Identity]], [[Set of all Self-Maps is Monoid]]
+}}
+{{eqn | r = x
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group}
+Tags: Symmetric Groups, Group Action of Symmetric Group
+
+\begin{theorem}
+Let $\N_n$ denote the [[Definition:Set|set]] $\set {1, 2, \ldots, n}$.
+Let $\struct {S_n, \circ}$ denote the [[Definition:Symmetric Group on n Letters|symmetric group]] on $\N_n$.
+The [[Definition:Mapping|mapping]] $*: S_n \times \N_n \to \N_n$ defined as:
+:$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \pi n$
+is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $\pi, \rho \in S_n$ and $n \in \N_n$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \pi * \paren {\rho * n}
+      | r = \pi * \map \rho n
+      | c = Definition of $*$
+}}
+{{eqn | r = \map \pi {\map \rho n}
+      | c = Definition of $*$
+}}
+{{eqn | r = \map {\paren {\pi \circ \rho} } n
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \paren {\pi \circ \rho} * n
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = I_{\N_n} * n
+      | r = \map {I_{\N_n} } n
+      | c = where $I_{\N_n}$ is the [[Definition:Identity Mapping|identity mapping]] on ${\N_n}$
+}}
+{{eqn | r = n
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group Acts Transitively}
+Tags: Group Action of Symmetric Group, Transitive Group Actions
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {\map \Gamma S, \circ}$ be the [[Definition:Symmetric Group|symmetric group]] on $S$.
+Let $*: \map \Gamma S \times S \to S$ be the [[Definition:Group Action|group action]] defined as:
+:$\forall \pi \in \map \Gamma S, \forall s \in S: \pi * s = \map \pi s$
+Then $*$ is a [[Definition:Transitive Group Action|transitive group action]].
+In other words, $\struct {\map \Gamma S, \circ}$ [[Definition:Transitive Group Action|acts transitively]] on $S$ by $*$.
+\end{theorem}
+
+\begin{proof}
+By [[Group Action of Symmetric Group]], $*: \map \Gamma S \times S \to S$ is indeed a [[Definition:Group Action|group action]]
+Let $s, t \in S$.
+As $\map \Gamma S$ is the [[Definition:Symmetric Group|symmetric group]] on $S$, there exists a [[Definition:Permutation|permutation]] $\pi \in \map \Gamma S$ such that:
+:$\map \pi s = t$
+This holds for any $s, t \in S$.
+Thus:
+:$\forall t \in S: t \in \Orb s$
+and so $S$ consists of a single [[Definition:Orbit (Group Theory)|orbit]].
+Hence the result by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Group Action is Group Action}
+Tags: Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $S$ be a [[Definition:Set|set]].
+Let $*: G \times S \to S$ be the [[Definition:Trivial Group Action|trivial group action]]:
+:$\forall \tuple {g, s} \in G \times S: g * s = s$
+Then $*$ is indeed a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $g_1, g_2 \in G$ and $s \in S$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = g_1 * \paren {g_2 * s}
+      | r = g_1 * s
+      | c = Definition of $*$
+}}
+{{eqn | r = s
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {g_1 \circ g_2} * s
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * s
+      | r = s
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+{{qed}}
+[[Category:Group Actions]]
+eutj6bymo8jbd3w84zt2vao3iax559s
+\end{proof}<|endoftext|>
+\section{Orbit of Trivial Group Action is Singleton}
+Tags: Trivial Group Action
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $S$ be a [[Definition:Set|set]].
+Let $*: G \times S \to S$ be the [[Definition:Trivial Group Action|trivial group action]]:
+:$\forall \left({g, s}\right) \in G \times S: g * s = s$
+Let $s \in S$.
+Then the [[Definition:Orbit (Group Theory)|orbit]] of $s$ under $*$ is $\left\{{s}\right\}$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\operatorname{Orb} \left({s}\right) = \left\{{t \in S: \exists g \in G: g * s = t}\right\}$
+By definition of the [[Definition:Trivial Group Action|trivial group action]]:
+:$\forall g \in G: g * s = s$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Regular Representation by Inverse is Group Action}
+Tags: Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Binary Operation|operation]]:
+:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
+where $\rho_g$ is the [[Definition:Right Regular Representation|right regular representation]] of $G$ with respect to $g$.
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $g, h, a \in G$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = g * \paren {h * a}
+      | r = g * \map {\rho_{h^{-1} } } a
+      | c = Definition of $*$
+}}
+{{eqn | r = g * \paren {a \circ h^{-1} }
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \map {\rho_{g^{-1} } } {a \circ h^{-1} }
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {a \circ h^{-1} } \circ g^{-1}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = a \circ \paren {h^{-1} \circ g^{-1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = a \circ \paren {g \circ h}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \map {\rho_{\paren {g \circ h}^{-1} } } a
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \paren {g \circ h} * a
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * s
+      | r = \map {\rho_{e^{-1} } } s
+      | c = Definition of $*$
+}}
+{{eqn | r = s \circ e^{-1}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = s \circ e
+      | c = [[Inverse of Identity Element is Itself]]
+}}
+{{eqn | r = s
+      | c = {{GroupAxiom|2}}
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Regular Representation is Group Action}
+Tags: Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Binary Operation|operation]]:
+:$\forall g, h \in G: g * h = \map {\lambda_g} h$
+where $\lambda_g$ is the [[Definition:Left Regular Representation|left regular representation]] of $G$ with respect to $g$.
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $g, h, a \in G$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = g * \paren {h * a}
+      | r = g * \map {\lambda_h} a
+      | c = Definition of $*$
+}}
+{{eqn | r = g * \paren {h \circ a}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \map {\lambda_g} {h \circ a}
+      | c = Definition of $*$
+}}
+{{eqn | r = g \circ \paren {h \circ a}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {g \circ h} \circ a
+      | c = [[Definition:Group Axioms|Group Axiom $G1$]]: [[Definition:Associative|associativity]]
+}}
+{{eqn | r = \map {\lambda_{\paren {g \circ h} } } a
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {g \circ h} * a
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * s
+      | r = \map {\lambda_e} s
+      | c = Definition of $*$
+}}
+{{eqn | r = e \circ s
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = s
+      | c = [[Definition:Group Axioms|Group Axiom $G2$]]: properties of [[Definition:Identity Element|identity element]]
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Regular Representation by Inverse is Transitive Group Action}
+Tags: Regular Representations, Transitive Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Group Action|group action]]:
+:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
+where $\rho_g$ is the [[Definition:Right Regular Representation|right regular representation]] of $G$ with respect to $g$.
+Then $*$ is a [[Definition:Transitive Group Action|transitive group action]].
+\end{theorem}
+
+\begin{proof}
+Let $g, h \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \exists a \in G:
+      | l = h
+      | r = a \circ g^{-1}
+      | c = [[Group has Latin Square Property]]
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = \map {\rho_{g^{-1} } } a
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = g * a
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+$h$ is arbitrary, therefore the above holds for all $h \in G$.
+By definition of [[Definition:Orbit (Group Theory)|orbit]]:
+:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
+That is:
+:$\Orb h = G$
+Hence the result by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Regular Representation is Transitive Group Action}
+Tags: Transitive Group Actions, Regular Representations
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Group Action|group action]]:
+:$\forall g, h \in G: g * h = \map {\lambda_g} h$
+where $\lambda_g$ is the [[Definition:Left Regular Representation|left regular representation]] of $G$ with respect to $g$.
+Then $*$ is a [[Definition:Transitive Group Action|transitive group action]].
+\end{theorem}
+
+\begin{proof}
+Let $g, h \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \exists a \in G:
+      | l = h
+      | r = g \circ a
+      | c = [[Group has Latin Square Property]]
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = \map {\lambda_g} a
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = g * a
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+$h$ is arbitrary, therefore the above holds for all $h \in G$.
+By definition of [[Definition:Orbit (Group Theory)|orbit]]:
+:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
+That is:
+:$\Orb h = G$
+Hence the result by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Action is not Transitive}
+Tags: Conjugacy Action, Transitive Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Conjugacy Action|conjugacy group action]]:
+: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
+Then $*$ is ''not'' a [[Definition:Transitive Group Action|transitive group action]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+For $G$ to be a [[Definition:Transitive Group Action|transitive group action]], the [[Definition:Orbit (Group Theory)|orbit]] of any [[Definition:Element|element]] of $G$ needs to be the whole of $G$.
+Take $h = e$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall g \in G:
+      | l = g * e
+      | r = g \circ e \circ g^{-1}
+      | c = 
+}}
+{{eqn | r = g \circ g^{-1}
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = e
+      | c = {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Orbit (Group Theory)|orbit]]:
+:$\Orb e = \set e$
+Only when $G$ is the [[Definition:Trivial Group|trivial group]], that is: $G = \set e$, does $\Orb e = G$.
+Hence the result by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Action on Abelian Group is Trivial}
+Tags: Conjugacy Action
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $*: G \times G \to G$ be the [[Definition:Conjugacy Action|conjugacy group action]]:
+: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
+Then $*$ is a [[Definition:Trivial Group Action|trivial group action]].
+\end{theorem}
+
+\begin{proof}
+For $G$ to be a [[Definition:Trivial Group Action|trivial group action]], the [[Definition:Orbit (Group Theory)|orbit]] of any [[Definition:Element|element]] of $G$ is a [[Definition:Singleton|singleton]] containing only that [[Definition:Element|element]].
+Take $h \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall g \in G:
+      | l = g * h
+      | r = g \circ h \circ g^{-1}
+      | c = 
+}}
+{{eqn | r = h \circ g \circ g^{-1}
+      | c = {{Defof|Abelian Group}}: $g$ [[Definition:Commute|commutes]] with $h$
+}}
+{{eqn | r = h \circ e
+      | c = [[Definition:Group Axioms|Group axiom $G3$]]: properties of [[Definition:Inverse Element|inverse element]]
+}}
+{{eqn | r = h
+      | c = [[Definition:Group Axioms|Group axiom $G2$]]: properties of [[Definition:Identity Element|identity element]]
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Orbit (Group Theory)|orbit]]:
+:$\Orb h = \set h$
+Hence the result by definition of [[Definition:Trivial Group Action|trivial group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Subgroup by Left Regular Representation}
+Tags: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Binary Operation|operation]] defined as:
+:$\forall \left({h, g}\right) \in H \times G: h * g = \lambda_h \left({g}\right)$
+where $\lambda_h \left({g}\right)$ is the [[Definition:Left Regular Representation|left regular representation]] of $g$ by $h$.
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $h_1, h_2 \in H$ and $g \in G$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = h_1 * \paren {h_2 * g}
+      | r = h_1 * \paren {\map {\lambda_{h_2} } g}
+      | c = Definition of $*$
+}}
+{{eqn | r = h_1 * \paren {h_2 \circ g}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \map {\lambda_{h_1} } {h_2 \circ g}
+      | c = Definition of $*$
+}}
+{{eqn | r = h_1 \circ \paren {h_2 \circ g}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {h_1 \circ h_2} \circ g
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = \map {\lambda_{h_1 \circ h_2} } g
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {h_1 \circ h_2} * g
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * g
+      | r = \map {\lambda_e} g
+      | c = Definition of $*$
+}}
+{{eqn | r = e \circ g
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = g
+      | c = {{GroupAxiom|2}}
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+The [[Definition:Group Action Axioms|group action axioms]] are thus seen to be fulfilled, and so $*$ is a [[Definition:Group Action|group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Subgroup by Right Regular Representation}
+Tags: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Binary Operation|operation]] defined as:
+:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
+where $\map {\rho_{h^{-1} } } g$ is the [[Definition:Right Regular Representation|right regular representation]] of $g$ by $h^{-1}$.
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $h_1, h_2 \in H$ and $g \in G$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = h_1 * \paren {h_2 * g}
+      | r = h_1 * \paren {\map {\rho_{h_2^{-1} } } g}
+      | c = Definition of $*$
+}}
+{{eqn | r = h_1 * \paren {g \circ h_2^{-1} }
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \map {\rho_{h_1^{-1} } } {h_2^{-1} \circ g}
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {g \circ h_2^{-1} } \circ h_1^{-1}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = g \circ \paren {h_2^{-1} \circ h_1^{-1} }
+      | c = [[Definition:Group Axioms|Group Axiom $G1$]]: [[Definition:Associative|Associativity]]
+}}
+{{eqn | r = g \circ \paren {h_1 \circ h_2}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \map {\rho_{\paren {h_1 \circ h_2}^{-1} } } g
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \paren {h_1 \circ h_2} * g
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * g
+      | r = \map {\rho_e} g
+      | c = Definition of $*$
+}}
+{{eqn | r = g \circ e^{-1}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = g \circ e
+      | c = [[Inverse of Identity Element is Itself]]
+}}
+{{eqn | r = g
+      | c = [[Definition:Group Axioms|Group Axiom $G2$]]: Properties of [[Definition:Identity Element|identity element]]
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+The [[Definition:Group Action Axioms|group action axioms]] are thus seen to be fulfilled, and so $*$ is a [[Definition:Group Action|group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Subgroup by Right Regular Representation is not Transitive}
+Tags: Transitive Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Group Action|group action]] defined as:
+:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
+where $\map {\rho_{h^{-1} } } g$ is the [[Definition:Right Regular Representation|right regular representation]] of $g$ by $h^{-1}$.
+Then $*$ is not [[Definition:Transitive Group Action|transitive]].
+\end{theorem}
+
+\begin{proof}
+From [[Group Action on Subgroup by Right Regular Representation]] it is established that $*$ is a [[Definition:Group Action|group action]].
+From [[Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset]]:
+:$\forall x \in G: \Orb x = H x$
+where $H x$ is the [[Definition:Right Coset|right coset]] of $H$ by $x$.
+From [[Right Coset Space forms Partition]] it is apparent that $H x \ne G$ unless $H = G$.
+Thus $\Orb x \ne G$ and the result follows by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset}
+Tags: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Group Action|group action]] defined as:
+:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
+where $\map {\rho_{h^{-1} } } g$ is the [[Definition:Right Regular Representation|right regular representation]] of $g$ by $h^{-1}$.
+Let $x \in G$.
+Then the [[Definition:Orbit (Group Theory)|orbit]] of $x$ under $*$ is given by:
+:$\forall x \in G: \Orb x = H x$
+where $H x$ is the [[Definition:Right Coset|right coset]] of $H$ by $x$.
+\end{theorem}<|endoftext|>
+\section{Left Coset by Identity}
+Tags: Coset by Identity
+
+\begin{theorem}
+Then:
+: $e H = H$
+where $e H$ is the [[Definition:Left Coset|left coset]] of $H$ by $e$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = e H
+      | r = \set {y \in G: \exists h \in H: y = e h}
+      | c = {{Defof|Left Coset}} of $H$ by $e$
+}}
+{{eqn | r = \set {y \in G: \exists h \in H: y = h}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \set {y \in G: y \in H}
+      | c = 
+}}
+{{eqn | r = H
+      | c = 
+}}
+{{end-eqn}}
+So $e H = H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Coset by Identity}
+Tags: Coset by Identity
+
+\begin{theorem}
+Then:
+: $H = H e$
+where $H e$ is the [[Definition:Right Coset|right coset]] of $H$ by $e$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = H e
+      | r = \set {x \in G: \exists h \in H: x = h e}
+      | c = {{Defof|Right Coset}} of $H$ by $e$
+}}
+{{eqn | r = \set {x \in G: \exists h \in H: x = h}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \set {x \in G: x \in H}
+      | c = 
+}}
+{{eqn | r = H
+      | c = 
+}}
+{{end-eqn}}
+So $H = H e$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inversion Mapping is Permutation}
+Tags: Inversion Mappings, Permutation Theory, Inversion Mapping is Permutation
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$.
+Then $\iota$ is a [[Definition:Permutation|permutation]] on $G$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Inversion Mapping|inversion mapping]] on $G$ is the [[Definition:Mapping|mapping]] $\iota: G \to G$ defined by:
+:$\forall g \in G: \map \iota g = g^{-1}$
+where $g^{-1}$ is the [[Definition:Inverse Element|inverse]] or $g$.
+By [[Inversion Mapping is Involution]], $\iota$ is an [[Definition:Involution (Mapping)|involution]]:
+:$\forall g \in G: \map \iota {\map \iota g} = g$
+The result follows from [[Involution is Permutation]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Proof of Surjection ===
+Let $a \in G$.
+By definition of $\iota$:
+:$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$
+By [[Inverse of Inverse]]:
+:$\left({a^{-1}}\right)^{-1} = a$
+Hence $a$ has a [[Definition:Preimage of Element under Mapping|preimage]].
+Since $a$ was arbitrary, $\iota$ is a [[Definition:Surjection|surjection]].
+=== Proof of Injection ===
+Suppose for some $a, b \in G$ that:
+:$\iota \left({a}\right) = \iota \left({b}\right)$
+Then by the definition of $\iota$:
+:$a^{-1} = b^{-1}$
+It follows from [[Inverse in Group is Unique]] that:
+:$a = b$
+Hence $\iota$ is an [[Definition:Injection|injection]].
+{{qed|lemma}}
+Hence by definition $\iota$ is a [[Definition:Bijection|bijection]].
+A [[Definition:Bijection|bijection]] from a [[Definition:Set|set]] to itself is by definition a [[Definition:Permutation|permutation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inversion Mapping is Mapping}
+Tags: Inversion Mappings
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
+Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$.
+Then $\iota$ is indeed a [[Definition:Mapping|mapping]].
+\end{theorem}
+
+\begin{proof}
+To show that $\iota$ is a [[Definition:Mapping|mapping]], it is sufficient to show that:
+:$\iota \left({a}\right) \ne \iota \left({b}\right) \implies a \ne b$:
+{{begin-eqn}}
+{{eqn | l = \iota \left({a}\right)
+      | o = \ne
+      | r = \iota \left({b}\right)
+      | c = by hypothesis
+}}
+{{eqn | ll= \implies
+      | l = a^{-1}
+      | o = \ne
+      | r = b^{-1}
+      | c = by definition of $\iota$
+}}
+{{eqn | ll= \implies
+      | l = a \circ a^{-1} \circ b
+      | o = \ne
+      | r = a \circ b^{-1} \circ b
+      | c = [[Cancellation Laws]] and [[Definition:Group Axioms|Group Axiom $G1$]]: [[Definition:Associative|Associativity]]
+}}
+{{eqn | ll= \implies
+      | l = e \circ b
+      | o = \ne
+      | r = a \circ e
+      | c = [[Definition:Group Axioms|Group Axiom $G3$]]: Properties of [[Definition:Inverse Element|Inverse Element]]
+}}
+{{eqn | ll= \implies
+      | l = b
+      | o = \ne
+      | r = a
+      | c = [[Definition:Group Axioms|Group Axiom $G2$]]: Properties of [[Definition:Identity Element|Identity Element]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Inversion Mappings]]
+joe5rd10ivc7k8jnxmv77ctakpp05ft
+\end{proof}<|endoftext|>
+\section{Cartesian Product of Group Actions}
+Tags: Group Actions, Cartesian Product
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be [[Definition:Group Action|group actions]].
+Then the [[Definition:Operation|operation]] $*: G \times \paren {S \times T} \to S \times T$ defined as:
+:$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$
+is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $g, h \in G$ and $s, t \in S$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = g * \tuple {h * \tuple {s, t} }
+      | r = g * \tuple {h *_S s, h *_T t}
+      | c = Definition of $*$
+}}
+{{eqn | r = \tuple {g *_S \tuple {h *_S s}, g *_T \tuple {h *_T t} }
+      | c = Definition of $*$
+}}
+{{eqn | r = \tuple {\paren {g \circ h} *_S s, \paren {g \circ h} *_T t}
+      | c = {{GroupActionAxiom|1}} for both $*_S$ and $*_T$
+}}
+{{eqn | r = \paren {g \circ h} * \tuple {s, t}
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * \tuple {s, t}
+      | r = \tuple {e *_S s, e *_T t}
+      | c = Definition of $*$
+}}
+{{eqn | r = \tuple {s, t}
+      | c = {{GroupActionAxiom|2}} for both $*_S$ and $*_T$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+The [[Definition:Group Action Axioms|group action axioms]] are thus seen to be fulfilled, and so $*$ is a [[Definition:Group Action|group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizer of Cartesian Product of Group Actions}
+Tags: Group Actions, Cartesian Product, Stabilizers
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be [[Definition:Group Action|group actions]].
+Let the [[Definition:Group Action|group action]] $*: G \times \paren {S \times T} \to S \times T$ be defined as:
+:$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$
+Then the [[Definition:Stabilizer|stabilizer]] of $\tuple {s, t} \in S \times T$ is given by:
+:$\Stab {s, t} = \Stab s \cap \Stab t$
+where $\Stab s$ and $\Stab t$ are the [[Definition:Stabilizer|stabilizers]] of $s$ and $t$ under $*_S$ and $*_T$ respectively.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Stabilizer|stabilizer]] of an [[Definition:Element|element]] $x$ of $S$ is defined as:
+:$\Stab x := \set {g \in G: g * x = x}$
+where $*$ denotes the [[Definition:Group Action|group action]].
+So:
+{{begin-eqn}}
+{{eqn | l = \Stab {s, t}
+      | r = \set {g \in G: g * \tuple {s, t} = \tuple {s, t} }
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {g \in G: \tuple {g *_S s, g *_T t} = \tuple {s, t} }
+      | c = Definition of $*$
+}}
+{{eqn | r = \set {g \in G: g *_S s = s \land g *_T t = t}
+      | c = {{Defof|Cartesian Product}}
+}}
+{{eqn | r = \set {g \in G: g \in \Stab s \land g \in \Stab t}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \Stab s \cap \Stab t
+      | c = {{Defof|Set Intersection}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Index in Subgroup}
+Tags: Subgroups, Index of Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H, K$ be [[Definition:Subgroup|subgroups]] of [[Definition:Finite Index|finite index]] of $G$.
+Then:
+:$\index H {H \cap K} \le \index G K$
+where $\index G K$ denotes the [[Definition:Index of Subgroup|index]] of $K$ in $G$.
+Equality happens {{iff}} $G = H K$.
+\end{theorem}
+
+\begin{proof}
+We list out all the [[Definition:Left Coset|left cosets]] of $H \cap K$ in $H$:
+:$H / \paren {H \cap K} = \set {h_n \paren {H \cap K}: h_n \in H, n \in I}$
+where $I$ is some [[Definition:Finite Set|finite]] indexing set.
+For each pair $h_i, h_j \in H \subseteq G$, where $i \ne j$:
+:$h_i^{-1} h_j \notin H \cap K \quad$ [[Cosets are Equal iff Product with Inverse in Subgroup]]
+Since $H$ is a [[Definition:Subgroup|subgroup]], $h_i^{-1} h_j \in H$.
+Thus:
+:$h_i^{-1} h_j \notin K$
+By [[Cosets are Equal iff Product with Inverse in Subgroup]], $h_i K$ and $h_j K$ are different [[Definition:Left Coset|left cosets]].
+Hence there must be at least $\index H {H \cap K}$ [[Definition:Left Coset|left cosets]] of $K$ in $G$, namely $h_n K$ for each $n \in I$.
+It follows that $\index H {H \cap K} \le \index G K$.
+{{qed|lemma}}
+=== Equality ===
+Suppose equality holds.
+Then the construction above gives all [[Definition:Left Coset|left cosets]] of $K$ in $G$.
+For any $x \in G$, $x \in h_i K$ for some $h_i \in H$.
+Hence $G \subseteq H K$.
+The inclusion $H K \subseteq G$ is seen from {{GroupAxiom|0}}.
+Hence $G = H K$ by [[Definition:Set Equality|definition of set equality]].
+{{qed|lemma}}
+Now suppose $G = H K$.
+To show equality we must show that the construction above gives all [[Definition:Left Coset|left cosets]] of $K$ in $G$.
+Let $g K$ be a [[Definition:Left Coset|left coset]] of $K$ in $G$, where $x \in G$.
+Since $G = H K$:
+:$\exists h \in H, k \in K: g = h k$
+Then by [[Left Coset Equals Subgroup iff Element in Subgroup]]:
+:$g K = h \paren {k K} = h K$
+Since $h \in H$:
+:$\exists i \in I: h \in h_i \paren {H \cap K}$
+By [[Element in Left Coset iff Product with Inverse in Subgroup]]:
+:$h^{-1} h_i \in H \cap K \subseteq K$
+Thus by [[Left Cosets are Equal iff Product with Inverse in Subgroup]]:
+:$h_i K = h K = g K$
+Hence all [[Definition:Left Coset|left cosets]] of $K$ in $G$ are obtained in the construction.
+This gives equality.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Cyclic Group has Euler Phi Generators}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $C_n$ be a [[Definition:Finite Cyclic Group|(finite) cyclic group]] of [[Definition:Order of Structure|order $n$]].
+Then $C_n$ has $\map \phi n$ [[Definition:Generator of Cyclic Group|generators]], where $\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+From [[List of Elements in Finite Cyclic Group]], the [[Definition:Element|elements]] of $G$ are:
+:$\set {g^k: g \in G, 0 \le k < n}$
+From [[Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order]], $g^k$ [[Definition:Generator of Cyclic Group|generates]] $G$ {{iff}} $k \perp n$.
+The result follows by definition of the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Left Regular Representations}
+Tags: Semigroups, Regular Representations
+
+\begin{theorem}
+:$\lambda_x \circ \lambda_y = \lambda_{x * y}$
+\end{theorem}
+
+\begin{proof}
+Let $z \in S$.
+{{begin-eqn}}
+{{eqn | l = \map {\lambda_x \circ \lambda_y} z
+      | r = \map {\lambda_x} {\map {\lambda_y} z}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\lambda_x} {y * z}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = x * \paren {y * z}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {x * y} * z
+      | c = {{SemigroupAxiom|1}}
+}}
+{{eqn | r = \map {\lambda_{x * y} } z
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Right Regular Representations}
+Tags: Semigroups, Regular Representations
+
+\begin{theorem}
+: $\rho_x \circ \rho_y = \rho_{y * x}$
+\end{theorem}
+
+\begin{proof}
+Let $z \in S$.
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\rho_x \circ \rho_y} } z
+      | r = \map {\rho_x} {\map {\rho_y} z}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\rho_x} {z * y}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \paren {z * y} * x
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = z * \paren {y * x}
+      | c = {{SemigroupAxiom|1}}
+}}
+{{eqn | r = \map {\rho_{y * x} } z
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Left Regular Representation with Right}
+Tags: Semigroups, Regular Representations
+
+\begin{theorem}
+: $\lambda_x \circ \rho_y = \rho_y \circ \lambda_x$
+\end{theorem}
+
+\begin{proof}
+Let $z \in S$.
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\lambda_x \circ \rho_y} } z
+      | r = \map {\lambda_x} {\map {\rho_y} z}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\lambda_x} {z * y}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = x * \paren {z * y}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \paren {x * z} * y
+      | c = {{SemigroupAxiom|1}}
+}}
+{{eqn | r = \map {\rho_y} {x * z}
+      | c = {{Defof|Right Regular Representation}}
+}}
+{{eqn | r = \map {\rho_y} {\map {\lambda_x} z}
+      | c = {{Defof|Left Regular Representation}}
+}}
+{{eqn | r = \map {\paren {\rho_y \circ \lambda_x} } z
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extendability Theorem for Derivatives Continuous on Open Intervals}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Continuous Real Function|continuous real function]] defined on an [[Definition:Real Interval|interval]] $\closedint a b$ where $a < b$.
+Then $f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\closedint a b$ {{iff}}:
+:$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint a b$
+and:
+:$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\closedint a b$.
+We need to show that:
+:$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint a b$
+and:
+:$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist.
+$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint a b$ because:
+:$f$ is continuously differentiable on $\closedint a b$
+:$\openint a b$ is a [[Definition:Subset|subset]] of $\closedint a b$. 
+What remains to show is that $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist.
+We shall only make a proof of the existence of $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ as the proof of the existence of $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ is similar.
+Since $f$ is [[Definition:Differentiable Real Function|differentiable]] at $a$, $\map {f'} a$ exists.
+Since $f'$ is [[Definition:Continuous Real Function|continuous]] at $a$, the following equation is true:
+:$\displaystyle \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$
+Therefore the {{RHS}} of this equation exists.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that:
+:$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint a b$
+and:
+:$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist.
+We need to show that $f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\closedint a b$.
+We know that $f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint a b$ so what remains to show is the following:
+:$\map {f'} a$ exists
+:$f'$ is [[Definition:Continuous Real Function|continuous]] at $a$
+:$\map {f'} b$ exists
+:$f'$ is [[Definition:Continuous Real Function|continuous]] at $b$.
+We shall limit ourselves to constructing a proof of the first two of these statements as the proof of the last two statements is similar.
+Since $a$ is the left-most [[Definition:Endpoint of Real Interval|end point]] of the [[Definition:Domain of Mapping|domain]] of $f$, $\map {f'} a$ is the [[Definition:Right-Hand Derivative|right-hand derivative]] of $f$ at $a$, denoted by $\map {f'_+} a$.
+We have per definition:
+:$\displaystyle \map {f'_+} a = \lim_{\delta \mathop \to 0^+} \frac {\map f {a + \delta} - \map f a} \delta$
+Since $f$ is [[Definition:Continuous Real Function|continuous]] on $\closedint a b$ and [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$, the [[Mean Value Theorem]] gives that a point $c$ in $\openint a  {a + \delta}$ exists such that
+:$\map f {a + \delta} - \map f a = \map {f'} c \delta$
+We use this equation in the expression for $\map {f'_+} a$:
+{{begin-eqn}}
+{{eqn | l = \map {f'_+} a
+      | r = \lim_{\delta \mathop \to 0^+} \frac {\map f {a + \delta} - \map f a} \delta
+}}
+{{eqn | r = \lim_{\delta \mathop \to 0^+} \frac {\map {f'} c \, \delta} \delta
+}}
+{{eqn | r = \lim_{\delta \mathop \to 0^+} \map {f'} c
+}}
+{{end-eqn}}
+Since $c$ is in $\openint a {a + \delta}$, we have $\size {c - a} < \delta$.
+Therefore, $c$ approaches $a$ when $\delta$ approaches 0.
+Moreover, since $c > a$, $c$ approaches $a$ from above.
+Since $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ exists, so does $\displaystyle \lim_{\delta \mathop \to 0^+} \map {f'} c$ because $c \to a^+$ when $\delta \to 0^+$.
+Furthermore, they are equal.
+Hence, since we proved above that $\displaystyle \lim_{\delta \mathop \to 0^+} \map {f'} c$ equals $\map {f'_+} a$
+:$\displaystyle \map {f'_+} a = \lim_{x \mathop \to a^+} \map {f'} x$
+Since $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ exists per presupposition, we conclude that $\map {f'_+} a$ exists, and that their values are equal.
+Hence $\map {f'} a$ exists and equals $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$.
+Thus, the first of the two statements we needed to prove is proven.
+Now we shall prove that $f'$ is [[Definition:Continuous Real Function|continuous]] at $a$.
+Since $a$ is the left-most [[Definition:Endpoint of Real Interval|end point]] of the [[Definition:Domain of Mapping|domain]] of $f$, $f'$ being [[Definition:Continuous Real Function|continuous]] at $a$ means that $\displaystyle \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$.
+Since we just proved this, we conclude that $f'$ is [[Definition:Continuous Real Function|continuous]] at $a$.
+{{qed}}
+[[Category:Differential Calculus]]
+iclshe4fmbqwbgj02hgacv3sdfkeiq5
+\end{proof}<|endoftext|>
+\section{Universal Affirmative and Universal Negative are Contrary iff First Predicate is not Vacuous}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\mathbf A: \quad$ The [[Definition:Universal Affirmative|universal affirmative]]: $\forall x: \map S x \implies \map P x$
+:$\mathbf E: \quad$ The [[Definition:Universal Negative|universal negative]]: $\forall x: \map S x \implies \neg \map P x$
+Then:
+:$\mathbf A$ and $\mathbf E$ are [[Definition:Contrary Statements|contrary]]
+{{iff}}:
+:$\exists x: \map S x$
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\exists x: \map S x \iff \neg \paren {\paren {\forall x: \map S x \implies \map P x} \land \paren {\forall x: \map S x \implies \neg \map P x} }$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\exists x: \map S x$.
+Suppose $\mathbf A$ and $\mathbf E$ are both [[Definition:True|true]].
+As $\mathbf A$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]:
+:$\map P x$
+As $\mathbf E$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]:
+:$\neg \map P x$
+It follows by [[Proof by Contradiction]] that $\mathbf A$ and $\mathbf E$ are not both [[Definition:True|true]].
+Thus, by definition, $\mathbf A$ and $\mathbf E$ are [[Definition:Contrary Statements|contrary statements]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\mathbf A$ and $\mathbf E$ be [[Definition:Contrary Statements|contrary statements]].
+Suppose:
+:$\neg \exists x: \map S x$
+that is, $\map S x$ is [[Definition:Vacuous Truth|vacuous]].
+From [[Denial of Existence]]:
+:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
+it follows that $\forall x: \map S x$ is [[Definition:False|false]].
+From [[False Statement implies Every Statement]]:
+:$\forall x: \map S x \implies \map P x$
+is [[Definition:True|true]], and:
+:$\forall x: \map S x \implies \neg \map P x$
+is also [[Definition:True|true]].
+Thus, by definition, $\mathbf A$ and $\mathbf E$ are not [[Definition:Contrary Statements|contrary statements]].
+It follows by [[Proof by Contradiction]] that $\exists x: \map S x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Affirmative and Particular Negative are Subcontrary iff First Predicate is not Vacuous}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\mathbf I: \quad$ The [[Definition:Particular Affirmative|particular affirmative]]: $\exists x: \map S x \land \map P x$
+:$\mathbf O: \quad$ The [[Definition:Particular Negative|particular negative]]: $\exists x: \map S x \land \neg \map P x$
+Then:
+:$\mathbf I$ and $\mathbf O$ are [[Definition:Subcontrary Statements|subcontrary]]
+{{iff}}:
+:$\exists x: \map S x$
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\exists x: \map S x \iff \paren {\paren {\exists x: \map S x \land \map P x} \lor \paren {\exists x: \map S x \land \neg \map P x} }$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\exists x: \map S x$.
+Suppose $\mathbf I$ and $\mathbf O$ are both [[Definition:False|false]].
+As $\mathbf I$ is [[Definition:False|false]], then by the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\neg \map P x$
+As $\mathbf O$ is [[Definition:False|false]], then by the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\neg \neg \map P x$
+and so by [[Double Negation]]:
+:$\map P x$
+It follows by [[Proof by Contradiction]] that $\mathbf I$ and $\mathbf O$ are not both [[Definition:False|false]].
+Thus, by definition, $\mathbf I$ and $\mathbf O$ are [[Definition:Subcontrary Statements|subcontrary statements]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\mathbf I$ and $\mathbf O$ be [[Definition:Subcontrary Statements|subcontrary statements]]
+Suppose $\neg \exists x: \map S x$.
+From the definition of [[Definition:Conjunction|logical conjunction]], it follows that:
+:$\neg \paren {\exists x: \map S x \land \map P x}$
+Similarly from the definition of [[Definition:Conjunction|logical conjunction]], it follows that:
+:$\neg \paren {\exists x: \map S x \land \neg \map P x}$
+That is, both $\mathbf I$ and $\mathbf O$ are [[Definition:False|false]].
+So, by definition, $\mathbf I$ and $\mathbf O$ are not [[Definition:Subcontrary Statements|subcontrary]].
+It follows by [[Proof by Contradiction]] that $\exists x: \map S x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Affirmative and Particular Negative are Contradictory}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\mathbf A: \quad$ The [[Definition:Universal Affirmative|universal affirmative]]: $\forall x: \map S x \implies \map P x$
+:$\mathbf O: \quad$ The [[Definition:Particular Negative|particular negative]]: $\exists x: \map S x \land \neg \map P x$
+Then $\mathbf A$ and $\mathbf O$ are [[Definition:Contradictory Statements|contradictory]].
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\neg \paren {\paren {\forall x: \map S x \implies \map P x} \iff \paren {\exists x: \map S x \land \neg \map P x} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \mathbf A
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \forall x:
+      | o = 
+      | r = \map S x \implies \map P x
+      | c = Definition of $\mathbf A$
+}}
+{{eqn | ll= \leadsto
+      | l = \forall x:
+      | o = 
+      | r = \neg \paren {\map S x \land \neg \map P x}
+      | c = [[Implication Equivalent to Negation of Conjunction with Negative]]
+}}
+{{eqn | ll= \leadsto
+      | l = \neg \exists x:
+      | o = 
+      | r = \map S x \land \neg \map P x
+      | c = [[De Morgan's Laws (Predicate Logic)/Denial of Existence|De Morgan's Laws: Denial of Existence]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \neg \mathbf O
+      | c = Definition of $\mathbf O$
+}}
+{{end-eqn}}
+The argument reverses:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \mathbf O
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x:
+      | o = 
+      | r = \map S x \land \neg \map P x
+      | c = Definition of $\mathbf O$
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x:
+      | o = 
+      | r = \neg \paren {\map S x \implies \map P x}
+      | c = [[Conjunction with Negative Equivalent to Negation of Implication]]
+}}
+{{eqn | ll= \leadsto
+      | l = \neg \forall x:
+      | o = 
+      | r = \map S x \implies \map P x
+      | c = [[De Morgan's Laws (Predicate Logic)/Denial of Universality|De Morgan's Laws: Denial of Universality]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \neg \mathbf A
+      | c = Definition of $\mathbf A$
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Contradictory Statements|contradictory]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Affirmative and Universal Negative are Contradictory}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\mathbf I: \quad$ The [[Definition:Particular Affirmative|particular affirmative]]: $\exists x: \map S x \land \map P x$
+:$\mathbf E: \quad$ The [[Definition:Universal Negative|universal negative]]: $\forall x: \map S x \implies \neg \map P x$
+Then $\mathbf I$ and $\mathbf E$ are [[Definition:Contradictory Statements|contradictory]].
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\neg \paren {\paren {\exists x: \map S x \land \map P x} \iff \paren {\forall x: \map S x \implies \neg \map P x} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \mathbf E
+      | c = 
+}}
+{{eqn | ll= \therefore
+      | l = \forall x:
+      | o = 
+      | r = \map S x \implies \neg \map P x
+      | c = Definition of $\mathbf E$
+}}
+{{eqn | ll= \therefore
+      | l = \forall x:
+      | o = 
+      | r = \neg \paren {\map S x \land \map P x}
+      | c = [[Modus Ponendo Tollens/Variant|Modus Ponendo Tollens]]
+}}
+{{eqn | ll= \therefore
+      | l = \neg \exists x:
+      | o = 
+      | r = \map S x \land \map P x
+      | c = [[De Morgan's Laws (Predicate Logic)/Denial of Existence|De Morgan's Laws: Denial of Existence]]
+}}
+{{eqn | ll= \therefore
+      | o = 
+      | r = \neg \mathbf I
+      | c = Definition of $\mathbf I$
+}}
+{{end-eqn}}
+The argument reverses:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \mathbf I
+      | c = 
+}}
+{{eqn | ll= \therefore
+      | l = \exists x:
+      | o = 
+      | r = \map S x \land \map P x
+      | c = Definition of $\mathbf I$
+}}
+{{eqn | ll= \therefore
+      | l = \exists x:
+      | o = 
+      | r = \neg \paren {\map S x \implies \neg \map P x}
+      | c = [[Conjunction Equivalent to Negation of Implication of Negative]]
+}}
+{{eqn | ll= \therefore
+      | l = \neg \forall x:
+      | o = 
+      | r = \map S x \implies \neg \map P x
+      | c = [[De Morgan's Laws (Predicate Logic)/Denial of Universality|De Morgan's Laws: Denial of Universality]]
+}}
+{{eqn | ll= \therefore
+      | o = 
+      | r = \neg \mathbf E
+      | c = Definition of $\mathbf E$
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Contradictory Statements|contradictory]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\map {\mathbf A} {S, P}: \quad$ The [[Definition:Universal Affirmative|universal affirmative]]: $\forall x: \map S x \implies \map P x$
+:$\map {\mathbf I} {S, P}: \quad$ The [[Definition:Particular Affirmative|particular affirmative]]: $\exists x: \map S x \land \map P x$
+Then:
+:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$
+{{iff}}:
+:$\exists x: \map S x$
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\exists x: \map S x$.
+Let $\map {\mathbf A} {S, P}$ be [[Definition:True|true]].
+As $\map {\mathbf A} {S, P}$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]:
+:$\map P x$
+From the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\map S x \land \map P x$
+Thus $\map {\mathbf I} {S, P}$ holds.
+So by the [[Rule of Implication]]:
+:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$.
+Suppose:
+:$\neg \exists x: \map S x$
+that is, $\map S x$ is [[Definition:Vacuous Truth|vacuous]].
+From [[De Morgan's Laws (Predicate Logic)/Denial of Existence|De Morgan's Laws: Denial of Existence]]:
+:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
+it follows that $\forall x: \map S x$ is [[Definition:False|false]].
+From [[False Statement implies Every Statement]]:
+:$\forall x: \map S x \implies \map P x$
+is [[Definition:True|true]].
+So $\map {\mathbf A} {S, P}$ holds.
+Again, $\neg \exists x: \map S x$.
+Then by the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\neg \paren {\exists x: \map S x \land \map P x}$
+That is, $\mathbf I$ does not hold.
+So $\map {\mathbf A} {S, P}$ is [[Definition:True|true]] and $\map {\mathbf I} {S, P}$ is [[Definition:False|false]].
+This [[Proof by Contradiction|contradicts]] $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$ by definition of [[Definition:Conditional|implication]].
+Thus $\exists x: \map S x$ must hold.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Negative implies Particular Negative iff First Predicate is not Vacuous}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\map {\mathbf E} {S, P}: \quad$ The [[Definition:Universal Negative|universal negative]]: $\forall x: \map S x \implies \neg \map P x$
+:$\map {\mathbf O} {S, P}: \quad$ The [[Definition:Particular Negative|particular negative]]: $\exists x: \map S x \land \neg \map P x$
+Then:
+:$\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$
+{{iff}}:
+:$\exists x: \map S x$
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \neg \map P x} \implies \paren {\exists x: \map S x \land \neg \map P x} }$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\exists x: \map S x$.
+Let $\map {\mathbf E} {S, P}$ be [[Definition:True|true]].
+As $\map {\mathbf E} {S, P}$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]:
+:$\neg \map P x$
+From the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\map S x \land \neg \map P x$
+Thus $\map {\mathbf O} {S, P}$ holds.
+So by the [[Rule of Implication]]:
+:$\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$.
+{{AimForCont}}:
+:$\neg \exists x: \map S x$
+that is, $\map S x$ is [[Definition:Vacuous Truth|vacuous]].
+From [[Denial of Existence|De Morgan's Laws: Denial of Existence]]:
+:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
+it follows that $\forall x: \map S x$ is [[Definition:False|false]].
+From [[False Statement implies Every Statement]]:
+:$\forall x: \map S x \implies \neg \map P x$
+is [[Definition:True|true]].
+So $\map {\mathbf E} {S, P}$ holds.
+Again, $\neg \exists x: \map S x$.
+Then by the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]:
+:$\neg \paren {\exists x: \map S x \land \neg \map P x}$
+That is, $\map {\mathbf O} {S, P}$ does not hold.
+So $\map {\mathbf E} {S, P}$ is [[Definition:True|true]] and $\map {\mathbf O} {S, P}$ is [[Definition:False|false]].
+This [[Proof by Contradiction|contradicts]] $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$ by definition of [[Definition:Conditional|implication]].
+Thus $\exists x: \map S x$ must hold.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Socrates is Mortal}
+Tags: Predicate Logic, Classic Problems
+
+\begin{theorem}
+:$(1): \quad$ ''All humans are mortal.''
+:$(2): \quad$ ''{{AuthorRef|Socrates}} is human.''
+:$(3): \quad$ ''Therefore {{AuthorRef|Socrates}} is mortal.''
+\end{theorem}
+
+\begin{proof}
+Let $x$ be an [[Definition:Object Variable|object variable]] from the [[Definition:Universe of Discourse|universe]] of '''rational beings'''.
+Let $\map H x$ denote the [[Definition:Propositional Function|propositional function]] ''$x$ is '''human'''''.
+Let $\map M x$ denote the [[Definition:Propositional Function|propositional function]] ''$x$ is '''mortal'''''.
+Let $S$ be a [[Definition:Proper Name|proper name]] that denotes {{AuthorRef|Socrates}}.
+The argument can then be expressed as:
+{{begin-eqn}}
+{{eqn | n = 1
+      | lo= \forall x:
+      | l = \map H x
+      | o = \implies
+      | r = \map M x
+      | c = 
+}}
+{{eqn | ll= \therefore
+      | l = \map H S
+      | o = \implies
+      | r = \map M S
+      | c = [[Universal Instantiation]]
+}}
+{{eqn | n = 2
+      | l = \map H S
+      | o = 
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \therefore
+      | l = \map M S
+      | o = 
+      | c = [[Modus Ponendo Ponens]]
+}}
+{{end-eqn}}
+That is: 
+:''{{AuthorRef|Socrates}} is mortal.''
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right-Hand Differentiable Function is Right-Continuous}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $a$ be a point in $I$ where $f$ is [[Definition:Right-Hand Derivative|right-hand differentiable]].
+Then $f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$.
+\end{theorem}
+
+\begin{proof}
+By hypothesis, $f'_+ \left({a}\right)$ exists.
+First we note that $a$ cannot be the right hand [[Definition:Endpoint of Real Interval|end point]] of $I$ because values in $I$ greater than $a$ need to exist for $f'_+ \left({a}\right)$ to exist.
+We form the following expression:
+:$\displaystyle \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)$
+We need to show that it is defined and to find its value.
+We find:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \lim_{x \mathop \to a^+} \left({\frac {f \left({x}\right) − f \left({a}\right)} {x − a} \left({x − a}\right)}\right)
+      | c = where the [[Definition:Denominator|denominator]] is unequal to $0$ since $x > a$
+}}
+{{eqn | r = \lim_{x \mathop \to a^+} \left({\frac {f \left({x}\right) − f \left({a}\right)} {x − a} }\right) \lim_{x \mathop \to a^+} \left({x−a}\right)
+      | c = [[Product Rule for Limits of Functions]] since (see the next step) the two limits exist
+}}
+{{eqn | r = f'_+ \left({a}\right) \times 0
+      | c = {{Defof|Right-Hand Derivative}}
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Note that this proves that $\displaystyle \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)$ exists.
+We continue by manipulating the result above:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)
+      | r = 0
+}}
+{{eqn | ll= \iff
+      | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right) + f \left({a}\right) - f \left({a}\right)
+      | r = 0
+}}
+{{eqn | ll= \iff
+      | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right) + \lim_{x \mathop \to a^+} f \left({a}\right) - f \left({a}\right)
+      | r = 0
+}}
+{{eqn | ll= \iff
+      | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right) + f \left({a}\right)}\right) - f \left({a}\right)
+      | r = 0
+      | c = [[Sum Rule for Limits of Functions]] since the two limits in the previous expression exist
+}}
+{{eqn | ll= \iff
+      | l = \lim_{x \mathop \to a^+} f \left({x}\right) - f \left({a}\right)
+      | r = 0
+}}
+{{eqn | ll= \iff
+      | l = \lim_{x \mathop \to a^+} f \left({x}\right)
+      | r = f \left({a}\right)
+}}
+{{end-eqn}}
+	
+which means that $f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$.
+{{qed}}
+[[Category:Continuous Functions]]
+9no786rkltfcbpsn4curh9qhem8d0wf
+\end{proof}<|endoftext|>
+\section{Left-Hand Differentiable Function is Left-Continuous}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $a$ be a point in $I$ where $f$ is [[Definition:Left-Hand Derivative|left-hand differentiable]].
+Then $f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $a$.
+\end{theorem}
+
+\begin{proof}
+By hypothesis, $\map {f'_-} a$ exists.
+First we note that $a$ cannot be the left-hand [[Definition:Endpoint of Real Interval|end point]] of $I$ because values in $I$ less than $a$ need to exist for $\map {f'_-} a$ to exist.
+We form the following expression:
+:$\displaystyle \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}$
+We need to show that it is defined and to find its value.
+We find:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}
+      | c = 
+}}
+{{eqn | r = \lim_{x \mathop \to a^-} \paren {\frac {\map f x − \map f a} {x − a} \paren {x − a} }
+      | c = where the [[Definition:Denominator|denominator]] is unequal to $0$ since $x < a$
+}}
+{{eqn | r = \lim_{x \mathop \to a^-} \paren {\frac {\map f x − \map f a} {x − a} } \lim_{x \mathop \to a^-} \paren {x − a}
+      | c = [[Product Rule for Limits of Functions]] since (see the next step) the two limits exist
+}}
+{{eqn | r = \map {f'_-} a \times 0
+      | c = {{Defof|Left-Hand Derivative}}
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Note that this proves that $\displaystyle \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}$ exists.
+We continue by manipulating the result above:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a} + \map f a - \map f a
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a} + \lim_{x \mathop \to a^-} \map f a - \map f a
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a + \map f a} - \map f a
+      | r = 0
+      | c = [[Sum Rule for Limits of Functions]] since the two limits in the previous expression exist
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{x \mathop \to a^-} \map f x - \map f a
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{x \mathop \to a^-} \map f x
+      | r = \map f a
+}}
+{{end-eqn}}
+	
+which means that $f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $a$.
+{{qed}}
+[[Category:Continuous Functions]]
+a5ozw10iwkagg95a83w0ncvfkxawznn
+\end{proof}<|endoftext|>
+\section{Left-Hand and Right-Hand Differentiable Function is Continuous}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $a$ be a point in $I$ where $f$ is [[Definition:Left-Hand Derivative|left-]] and [[Definition:Right-Hand Derivative|right-hand differentiable]].
+Then $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $a$.
+\end{theorem}
+
+\begin{proof}
+By [[Left-Hand Differentiable Function is Left-Continuous]], $f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $a$.
+By [[Right-Hand Differentiable Function is Right-Continuous]], $f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$.
+By [[Limit iff Limits from Left and Right]], $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $a$ since $f$ is [[Definition:Left-Continuous at Point|left-]] and [[Definition:Right-Continuous at Point|right-continuous]] at $a$.
+{{qed}}
+[[Category:Continuous Functions]]
+jlvd6r0vqqj5okpdfr9jep3h7cqz3k6
+\end{proof}<|endoftext|>
+\section{Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+{{begin-axiom}}
+{{axiom | q = \mathbf A \left({S, P}\right)
+        | lc= The [[Definition:Universal Affirmative|universal affirmative]]:
+        | ml= \forall x: S \left({x}\right)
+        | mo= \implies
+        | mr= P \left({x}\right)
+}}
+{{axiom | q = \mathbf E \left({S, P}\right)
+        | lc= The [[Definition:Universal Negative|universal negative]]:
+        | ml= \forall x: S \left({x}\right)
+        | mo= \implies
+        | mr= \neg P \left({x}\right)
+}}
+{{axiom | q = \mathbf I \left({S, P}\right)
+        | lc= The [[Definition:Particular Affirmative|particular affirmative]]:
+        | ml= \exists x: S \left({x}\right)
+        | mo= \land 
+        | mr= P \left({x}\right)
+}}
+{{axiom | q = \mathbf O \left({S, P}\right)
+        | lc= The [[Definition:Particular Negative|particular negative]]:
+        | ml= \exists x: S \left({x}\right)
+        | mo= \land 
+        | mr= \neg P \left({x}\right)
+}}
+{{end-axiom}}
+Then:
+:$\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ are both [[Definition:False|false]]
+{{iff}}:
+:$\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ are both [[Definition:True|true]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ both be [[Definition:False|false]].
+{{begin-eqn}}
+{{eqn | n = 1
+      | o =
+      | r = \neg \mathbf A \left({S, P}\right) \land \neg \mathbf E \left({S, P}\right)
+}}
+{{eqn | n = 2
+      | o = \leadsto
+      | r = \neg \mathbf A \left({S, P}\right)
+      | c = [[Rule of Simplification]]: from $(1)$
+}}
+{{eqn | n = 3
+      | o = \leadsto
+      | r = \mathbf O \left({S, P}\right)
+      | c = [[Universal Affirmative and Particular Negative are Contradictory]]: from $(2)$
+}}
+{{eqn | n = 4
+      | o = \leadsto
+      | r = \neg \mathbf E \left({S, P}\right)
+      | c = [[Rule of Simplification]]: from $(1)$
+}}
+{{eqn | n = 5
+      | o = \leadsto
+      | r = \mathbf I \left({S, P}\right)
+      | c = [[Particular Affirmative and Universal Negative are Contradictory]]: from $(4)$
+}}
+{{eqn | n = 6
+      | o = \leadsto
+      | r = \mathbf I \left({S, P}\right) \land \mathbf O \left({S, P}\right)
+      | c = [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]: from $(5)$ and $(3)$
+}}
+{{end-eqn}}
+Hence $\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ are both [[Definition:True|true]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ both be [[Definition:True|true]].
+{{begin-eqn}}
+{{eqn | n = 1
+      | o =
+      | r = \mathbf I \left({S, P}\right) \land \mathbf O \left({S, P}\right)
+}}
+{{eqn | n = 2
+      | o = \leadsto
+      | r = \mathbf I \left({S, P}\right)
+      | c = [[Rule of Simplification]]: from $(1)$
+}}
+{{eqn | n = 3
+      | o = \leadsto
+      | r = \neg \mathbf E \left({S, P}\right)
+      | c = [[Particular Affirmative and Universal Negative are Contradictory]]: from $(2)$
+}}
+{{eqn | n = 4
+      | o = \leadsto
+      | r = \mathbf O \left({S, P}\right)
+      | c = [[Rule of Simplification]]: from $(1)$
+}}
+{{eqn | n = 5
+      | o = \leadsto
+      | r = \neg \mathbf A \left({S, P}\right)
+      | c = [[Universal Affirmative and Particular Negative are Contradictory]]: from $(4)$
+}}
+{{eqn | n = 6
+      | o = \leadsto
+      | r = \neg \mathbf A \left({S, P}\right) \land \neg \mathbf E \left({S, P}\right)
+      | c = [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]: from $(5)$ and $(3)$
+}}
+{{end-eqn}}
+Hence $\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ are both [[Definition:False|false]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Simple Conversion of I}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Particular Affirmative|particular affirmative]] [[Definition:Categorical Statement|categorical statement]] ''Some $S$ is $P$'':
+:$\map {\mathbf I} {S, P}: \exists x: \map S x \land  \map P x$
+Then ''Some $P$ is $S$'':
+:$\map {\mathbf I} {P, S}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \map {\mathbf I} {S, P}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \exists x: \map S x \land \map P x
+      | c = {{Defof|Particular Affirmative}}
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \exists x: \map P x \land \map S x
+      | c = [[Conjunction is Commutative]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \map {\mathbf I} {P, S}
+      | c = {{Defof|Particular Affirmative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Simple Conversion of E}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Universal Negative|universal negative]] [[Definition:Categorical Statement|categorical statement]] ''No $S$ is $P$'':
+:$\mathbf E \left({S, P}\right): \forall x: S \left({x}\right) \implies \neg P \left({x}\right)$
+Then ''No $P$ is $S$'':
+:$\mathbf E \left({P, S}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \mathbf E \left({S, P}\right)
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = \forall x: S \left({x}\right) \implies \neg P \left({x}\right)
+      | c = Definition of [[Definition:Universal Negative|Universal Negative]]
+}}
+{{eqn | o = \implies
+      | r = \forall x: \neg \left({S \left({x}\right) \land P \left({x}\right)}\right)
+      | c = [[Modus Ponendo Tollens/Variant|Modus Ponendo Tollens]]
+}}
+{{eqn | o = \implies
+      | r = \forall x: \neg \left({P \left({x}\right) \land S \left({x}\right)}\right)
+      | c = [[Conjunction is Commutative]]
+}}
+{{eqn | o = \implies
+      | r = \forall x: P \left({x}\right) \implies \neg S \left({x}\right)
+      | c = [[Modus Ponendo Tollens/Variant|Modus Ponendo Tollens]]
+}}
+{{eqn | o = \implies
+      | r = \mathbf E \left({P, S}\right)
+      | c = Definition of [[Definition:Universal Negative|Universal Negative]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conversion per Accidens}
+Tags: Categorical Statements
+
+\begin{theorem}
+Consider the [[Definition:Categorical Statement|categorical statements]]:
+:$\map {\mathbf A} {S, P}: \quad$ The [[Definition:Universal Affirmative|universal affirmative]]: $\forall x: \map S x \implies \map P x$
+:$\map {\mathbf I} {P, S}: \quad$ The [[Definition:Particular Affirmative|particular affirmative]]: $\exists x: \map P x \land \map S x$
+Then:
+:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {P, S}$
+{{iff}}:
+:$\exists x: \map S x$
+Using the [[Definition:Symbolic Logic|symbology]] of [[Definition:Predicate Logic|predicate logic:]]
+:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map P x \land \map S x} }$
+This law has the traditional name '''conversion per accidens of $\mathbf A$'''.
+Thus the $\mathbf A$ form '''converts per accidens''' to the $\mathbf I$ form.
+\end{theorem}
+
+\begin{proof}
+From [[Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous]]:
+:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$
+From [[Law of Simple Conversion of I]]:
+:$\paren {\exists x: \map S x \land \map P x} \implies \paren {\exists x: \map P x \land \map S x}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Standard Instances of Categorical Syllogism}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+There are $256$ [[Definition:Distinct Elements|distinct]] [[Definition:Standard Instance of Categorical Syllogism|standard instances]] of the [[Definition:Categorical Syllogism|categorical syllogism]].
+\end{theorem}
+
+\begin{proof}
+Recall the four [[Definition:Figure of Categorical Syllogism|figures]] of the [[Definition:Categorical Syllogism|categorical syllogism]]:
+:$\begin{array}{r|rl}
+\text I & & \\
+\hline \\
+\text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\
+\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\
+\hline \\
+\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
+\end{array}
+\qquad
+\begin{array}{r|rl}
+\text {II} & & \\
+\hline \\
+\text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\
+\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\
+\hline \\
+\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
+\end{array}$
+:$\begin{array}{r|rl}
+\text {III} & & \\
+\hline \\
+\text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\
+\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\
+\hline \\
+\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
+\end{array}
+\qquad
+\begin{array}{r|rl}
+\text {IV} & & \\
+\hline \\
+\text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\
+\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\
+\hline \\
+\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
+\end{array}$
+By the definition of a [[Definition:Categorical Syllogism|categorical syllogism]], the following are fixed:
+:$(1): \quad$ The order of the [[Definition:Categorical Statement|categorical statements]]: [[Definition:Major Premise of Syllogism|major premise]], then [[Definition:Minor Premise of Syllogism|minor premise]], then [[Definition:Conclusion of Syllogism|conclusion]].
+:$(2): \quad$ The structure of the [[Definition:Conclusion of Syllogism|conclusion]]: the [[Definition:Secondary Term of Syllogism|secondary term]] then the [[Definition:Primary Term of Syllogism|primary term]].
+:$(3): \quad$ The content of the two [[Definition:Premise|premises]]: the [[Definition:Major Premise of Syllogism|major premise]] contains the [[Definition:Primary Term of Syllogism|primary term]] and the [[Definition:Middle Term of Syllogism|middle term]], while the [[Definition:Minor Premise of Syllogism|minor premise]] contains the [[Definition:Middle Term of Syllogism|middle term]] and the [[Definition:Secondary Term of Syllogism|secondary term]].
+The things that can be varied are:
+:$(A): \quad$ The specific nature of each of the [[Definition:Categorical Statement|categorical statements]]: each can be any of four types: $\mathbf A$, $\mathbf E$, $\mathbf I$ or $\mathbf O$.
+:$(B): \quad$ Within the [[Definition:Major Premise of Syllogism|major]] and [[Definition:Minor Premise of Syllogism|minor premises]], the order of the two [[Definition:Term of Syllogism|terms]]: the [[Definition:Middle Term of Syllogism|middle term]] can come either first or second.
+The order of the [[Definition:Term of Syllogism|terms]] within the [[Definition:Premise of Syllogism|premises]] determines the [[Definition:Figure of Categorical Syllogism|figure]] of the [[Definition:Categorical Syllogism|categorical syllogism]].
+For each of the two [[Definition:Premise of Syllogism|premises]] there are two options for the position of the [[Definition:Middle Term of Syllogism|middle term]].
+By the [[Product Rule for Counting]] it follows that there are $2 \times 2 = 4$ [[Definition:Figure of Categorical Syllogism|figures]].
+In each [[Definition:Figure of Categorical Syllogism|figure]], there are $3$ [[Definition:Categorical Statement|categorical statements]].
+Each can be of any of $4$ types.
+Hence by the [[Product Rule for Counting]], there are $4 \times 4 \times 4 = 64$ possible [[Definition:Standard Instance of Categorical Syllogism|standard instances]] for each [[Definition:Figure of Categorical Syllogism|figure]].
+For all $4$ [[Definition:Figure of Categorical Syllogism|figures]], by the [[Product Rule for Counting]], it follows that there are $4 \times 64 = 256$ [[Definition:Standard Instance of Categorical Syllogism|standard instances]] in total.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{No Valid Categorical Syllogism contains two Negative Premises}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+No [[Definition:Categorical Syllogism|categorical syllogism]] of which both [[Definition:Premise of Syllogism|premises]] are [[Definition:Negative Categorical Statement|negative categorical statements]] is [[Definition:Valid Argument|valid]].
+\end{theorem}<|endoftext|>
+\section{Valid Patterns of Categorical Syllogism}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+The following [[Definition:Categorical Syllogism|categorical syllogisms]] are valid:
+:$\begin{array}{rl}
+\text{I} & AAA \\
+\text{I} & AII \\
+\text{I} & EAE \\
+\text{I} & EIO \\
+* \text{I} & AAI \\
+* \text{I} & EAO \\
+\end{array}
+\qquad
+\begin{array}{rl}
+\text{II} & EAE \\
+\text{II} & AEE \\
+\text{II} & AOO \\
+\text{II} & EIO \\
+* \text{II} & EAO \\
+* \text{II} & AEO \\
+\end{array}
+\qquad
+\begin{array}{rl}
+\dagger \text{III} & AAI \\
+\text{III} & AII \\
+\text{III} & IAI \\
+\dagger \text{III} & EAO \\
+\text{III} & EIO \\
+\text{III} & OAO \\
+\end{array}
+\qquad
+\begin{array}{rl}
+\S \text{IV} & AAI \\
+\text{IV} & AEE \\
+\dagger \text{IV} & EAO \\
+\text{IV} & EIO \\
+\text{IV} & IAI \\
+* \text{IV} & AEO \\
+\end{array}$
+In the above:
+:$\text{I}, \text{II}, \text{III}, \text{IV}$ denote the four [[Definition:Figure of Categorical Syllogism|figures]] of the [[Definition:Categorical Syllogism|categorical syllogisms]]
+:$A, E, I, O$ denote the [[Definition:Universal Affirmative|universal affirmative]], [[Definition:Universal Negative|universal negative]], [[Definition:Particular Affirmative|particular affirmative]] and [[Definition:Particular Negative|particular negative]] respectively: see [[Definition:Shorthand for Categorical Syllogism|Shorthand for Categorical Syllogism]]
+:[[Definition:Categorical Syllogism|Syllogisms]] marked $*$ require the assumption that $\exists x: \map S x$, that is, that there exists an object fulfilling the [[Definition:Secondary Term of Syllogism|secondary predicate]]
+:[[Definition:Categorical Syllogism|Syllogisms]] marked $\dagger$ require the assumption that $\exists x: \map M x$, that is, that there exists an object fulfilling the [[Definition:Middle Term of Syllogism|middle predicate]]
+:[[Definition:Categorical Syllogism|Syllogisms]] marked $\S$ require the assumption that $\exists x: \map P x$, that is, that there exists an object fulfilling the [[Definition:Primary Term of Syllogism|primary predicate]]
+\end{theorem}
+
+\begin{proof}
+From [[Elimination of all but 24 Categorical Syllogisms as Invalid]], all but these $24$ patterns have been shown to be [[Definition:Invalid Argument|invalid]].
+It remains to be shown that these remaining syllogisms are in fact [[Definition:Valid Argument|valid]].
+{{ProofWanted|Considerable work to be done yet.}}
+\end{proof}<|endoftext|>
+\section{Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+Let $T$ be a [[Definition:Term of Syllogism|term]] of the [[Definition:Conclusion of Syllogism|conclusion]] $C$ of a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]] $Q$.
+Let $T$ be [[Definition:Distributed Term of Categorical Syllogism|distributed]] in $C$.
+Then $T$ is also [[Definition:Distributed Term of Categorical Syllogism|distributed]] in whichever [[Definition:Premise of Syllogism|premise]] of $Q$ in which it appears.
+\end{theorem}<|endoftext|>
+\section{Middle Term of Valid Categorical Syllogism is Distributed at least Once}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+Let $M$ be the [[Definition:Middle Term of Syllogism|middle term]] of a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]] $Q$.
+Then $M$ is [[Definition:Distributed Term of Categorical Syllogism|distributed]] in at least one of the [[Definition:Premise of Syllogism|premises]] of $Q$ in which it appears.
+\end{theorem}<|endoftext|>
+\section{Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+The [[Definition:Conclusion of Syllogism|conclusion]] of a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]] is [[Definition:Negative Categorical Statement|negative]] {{iff}} one of the [[Definition:Premise of Syllogism|premises]] is also [[Definition:Negative Categorical Statement|negative]].
+\end{theorem}<|endoftext|>
+\section{No Valid Categorical Syllogism contains two Particular Premises}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+Let $Q$ be a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]].
+Then at least one of the [[Definition:Premise of Syllogism|premises]] of $Q$ is [[Definition:Universal Categorical Statement|universal]].
+\end{theorem}
+
+\begin{proof}
+Suppose both [[Definition:Premise of Syllogism|premises]] of $Q$ are [[Definition:Particular Categorical Statement|particular]].
+Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the [[Definition:Conclusion of Syllogism|conclusion]].
+$\text{I}$ is neither [[Definition:Universal Categorical Statement|universal]] nor [[Definition:Negative Categorical Statement|negative]].
+Thus the $\text{II}x$ pattern does not [[Definition:Distributed Term of Categorical Syllogism|distribute]] the [[Definition:Middle Term of Syllogism|middle term]] of $Q$.
+So $\text{II}x$ violates the rule [[Middle Term of Valid Categorical Syllogism is Distributed at least Once]].
+$\text{O}$ is [[Definition:Negative Categorical Statement|negative]].
+Thus $\text{OO}x$ violates the rule [[No Valid Categorical Syllogism contains two Negative Premises]].
+It remains to investigate $\text{IO}x$ and $\text{OI}x$.
+By [[Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative]], the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ is [[Definition:Negative Categorical Statement|negative]], either $\text{E}$ or $\text{O}$.
+By the definition of [[Definition:Distributed Predicate of Categorical Syllogism|Distributed Predicate of Categorical Syllogism]], the [[Definition:Predicate of Categorical Statement|predicate]] of the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ is [[Definition:Distributed Term of Categorical Syllogism|distributed]].
+Thus, by construction, the [[Definition:Primary Term of Syllogism|primary term]] $P$ of $Q$ is [[Definition:Distributed Term of Categorical Syllogism|distributed]].
+By [[Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise]], it follows that $P$ is [[Definition:Distributed Term of Categorical Syllogism|distributed]] in the [[Definition:Major Premise of Syllogism|major premise]] of $Q$.
+This eliminates $\text{IO}x$ as the [[Definition:Particular Affirmative|particular affirmative]] $\text{I}$ [[Definition:Distributed Term of Categorical Syllogism|distributes]] neither its [[Definition:Subject of Categorical Statement|subject]] nor its [[Definition:Predicate of Categorical Statement|predicate]].
+The final pattern to be investigated is $\text{OI}x$.
+By definition, $\text{O}$ [[Definition:Distributed Term of Categorical Syllogism|distributes]] only its [[Definition:Predicate of Categorical Statement|predicate]].
+As $\text{O}$ needs to [[Definition:Distributed Term of Categorical Syllogism|distribute]] the [[Definition:Primary Term of Syllogism|primary term]] $P$, it cannot also [[Definition:Distributed Term of Categorical Syllogism|distribute]] the [[Definition:Middle Term of Syllogism|middle term]] $M$ of $Q$.
+But the [[Definition:Particular Affirmative|particular affirmative]] $\text{I}$ (as has been seen above) also does not [[Definition:Distributed Term of Categorical Syllogism|distribute]] $M$.
+Thus $M$ remains [[Definition:Undistributed Term of Categorical Syllogism|undistributed]], so violating the rule [[Middle Term of Valid Categorical Syllogism is Distributed at least Once]].
+This eliminates $\text{OI}x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{No Valid Categorical Syllogism with Particular Premise has Universal Conclusion}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+Let $Q$ be a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]].
+Let one of the [[Definition:Premise of Syllogism|premises]] of $Q$ be [[Definition:Particular Categorical Statement|particular]].
+Then the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ is also [[Definition:Particular Categorical Statement|particular]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Major Premise of Syllogism|major premise]] of $Q$ be denoted $\text{Maj}$.
+Let the [[Definition:Minor Premise of Syllogism|minor premise]] of $Q$ be denoted $\text{Min}$.
+Let the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ be denoted $\text{C}$.
+From [[No Valid Categorical Syllogism contains two Particular Premises]], either $\text{Maj}$ or $\text{Min}$ has to be [[Definition:Universal Categorical Statement|universal]].
+Let the other [[Definition:Premise of Syllogism|premise]] of $Q$ be [[Definition:Particular Categorical Statement|particular]].
+Suppose $\text{C}$ is the [[Definition:Universal Affirmative|universal affirmative]] $\mathbf A \left({S, P}\right)$.
+From [[Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative]] it follows that $\text{Maj}$ and $\text{Min}$ are also both [[Definition:Affirmative Categorical Statement|affirmative]].
+The only patterns fitting the criteria are $\text{AI}x$ and $\text{IA}x$, both of which [[Definition:Distributed Term of Categorical Syllogism|distribute]] only one [[Definition:Term of Syllogism|term]] in $Q$.
+We have that $\text{C}$ [[Definition:Distributed Term of Categorical Syllogism|distributes]] $S$.
+From [[Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise]], $S$ must also be [[Definition:Distributed Term of Categorical Syllogism|distributed]] in $\text{Min}$.
+From [[Middle Term of Valid Categorical Syllogism is Distributed at least Once]], the [[Definition:Middle Term of Syllogism|middle term]] $M$ of $Q$ needs to be [[Definition:Distributed Term of Categorical Syllogism|distributed]].
+Thus we require two [[Definition:Distributed Term of Categorical Syllogism|distributed terms]] in the [[Definition:Premise of Syllogism|premises]] for $\text{C}$ to be the [[Definition:Universal Affirmative|universal affirmative]].
+So if $\text{C}$ is [[Definition:Affirmative Categorical Statement|affirmative]], neither [[Definition:Premise of Syllogism|premise]] of $Q$ can be [[Definition:Particular Categorical Statement|particular]].
+From this contradiction it follows that $\text{C}$ is not $\mathbf A \left({S, P}\right)$.
+Suppose $\text{C}$ is the [[Definition:Universal Negative|universal negative]] $\mathbf E \left({S, P}\right)$.
+In this case, both $S$ and $P$ are [[Definition:Distributed Term of Categorical Syllogism|distributed]] in $C$, as the [[Definition:Universal Negative|universal negative]] [[Definition:Distributed Term of Categorical Syllogism|distributes]] both $S$ and $P$.
+	
+By [[Middle Term of Valid Categorical Syllogism is Distributed at least Once]], the [[Definition:Middle Term of Syllogism|middle term]] $M$ of $Q$ needs to be [[Definition:Distributed Term of Categorical Syllogism|distributed]].
+	
+Thus there are three terms needing to be [[Definition:Distributed Term of Categorical Syllogism|distributed]] in two [[Definition:Premise of Syllogism|premises]].
+So at least one [[Definition:Premise of Syllogism|premises]] needs to be the [[Definition:Universal Negative|universal negative]].
+The other premise is not $\mathbf O$ as this violates [[No Valid Categorical Syllogism contains two Negative Premises]].
+But it also is not $\mathbf I$, as $\mathbf I$ does not distribute any terms.
+So if $\text{C}$ is the [[Definition:Universal Negative|universal negative]], neither [[Definition:Premise of Syllogism|premise]] can be [[Definition:Particular Categorical Statement|particular]].
+From this contradiction it follows that $\text{C}$ is not $\mathbf E \left({S, P}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elimination of all but 48 Categorical Syllogisms as Invalid}
+Tags: Categorical Syllogisms
+
+\begin{theorem}
+Of the $256$ different types of [[Definition:Categorical Syllogism|categorical syllogism]], all but $48$ can immediately be identified as [[Definition:Invalid Argument|invalid]] by consideration of the [[Rules of Quantity]] and the [[Rules of Quality]].
+\end{theorem}
+
+\begin{proof}
+There are $64$ patterns of [[Definition:Categorical Syllogism|categorical syllogism]] per [[Definition:Figure of Categorical Syllogism|figure]]:
+:$\begin{array}{cccc}
+AAA & AAE & AAI & AAO \\
+AEA & AEE & AEI & AEO \\
+AIA & AIE & AII & AIO \\
+AOA & AOE & AOI & AOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+EAA & EAE & EAI & EAO \\
+EEA & EEE & EEI & EEO \\
+EIA & EIE & EII & EIO \\
+EOA & EOE & EOI & EOO \\
+\end{array}$
+:$\begin{array}{cccc}
+IAA & IAE & IAI & IAO \\
+IEA & IEE & IEI & IEO \\
+IIA & IIE & III & IIO \\
+IOA & IOE & IOI & IOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+OAA & OAE & OAI & OAO \\
+OEA & OEE & OEI & OEO \\
+OIA & OIE & OII & OIO \\
+OOA & OOE & OOI & OOO \\
+\end{array}$
+From [[No Valid Categorical Syllogism contains two Negative Premises]], all those whose patterns start $EE$, $EO$, $OE$ and $OO$ can be eliminated:
+:$\begin{array}{cccc}
+AAA & AAE & AAI & AAO \\
+AEA & AEE & AEI & AEO \\
+AIA & AIE & AII & AIO \\
+AOA & AOE & AOI & AOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+EAA & EAE & EAI & EAO \\
+  &  &  &  \\
+EIA & EIE & EII & EIO \\
+ &  &   &   \\
+\end{array}$
+:$\begin{array}{cccc}
+IAA & IAE & IAI & IAO \\
+IEA & IEE & IEI & IEO \\
+IIA & IIE & III & IIO \\
+IOA & IOE & IOI & IOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+OAA & OAE & OAI & OAO \\
+  &   &   &   \\
+OIA & OIE & OII & OIO \\
+  &   &  &   \\
+\end{array}$
+From [[No Valid Categorical Syllogism contains two Particular Premises]], all those whose patterns start $II$, $IO$ and $OI$ and $OO$ can be eliminated:
+:$\begin{array}{cccc}
+AAA & AAE & AAI & AAO \\
+AEA & AEE & AEI & AEO \\
+AIA & AIE & AII & AIO \\
+AOA & AOE & AOI & AOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+EAA & EAE & EAI & EAO \\
+  &  &  &  \\
+EIA & EIE & EII & EIO \\
+\end{array}$
+:$\begin{array}{cccc}
+IAA & IAE & IAI & IAO \\
+IEA & IEE & IEI & IEO \\
+\end{array} \qquad
+\begin{array}{cccc}
+OAA & OAE & OAI & OAO \\
+\end{array}$
+From [[Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative]], further patterns can be eliminated:
+:$\begin{array}{cccc}
+AAA & & AAI & \\
+ & AEE & & AEO \\
+AIA & & AII & \\
+ & AOE & & AOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+ & EAE & & EAO \\
+ &  &  &  \\
+ & EIE & & EIO \\
+\end{array}$
+:$\begin{array}{cccc}
+IAA & & IAI & \\
+ & IEE & & IEO \\
+\end{array} \qquad
+\begin{array}{cccc}
+ & OAE & & OAO \\
+\end{array}$
+From [[No Valid Categorical Syllogism with Particular Premise has Universal Conclusion]], all those whose patterns match that condition can be eliminated:
+:$\begin{array}{cccc}
+AAA & & AAI & \\
+ & AEE & & AEO \\
+ & & AII & \\
+ & & & AOO \\
+\end{array} \qquad
+\begin{array}{cccc}
+ & EAE & & EAO \\
+ &  &  &  \\
+ & & & EIO \\
+\end{array}$
+:$\begin{array}{cccc}
+ & & IAI & \\
+ & & & IEO \\
+\end{array} \qquad
+\begin{array}{cccc}
+ & & & OAO \\
+\end{array}$
+Thus there are $12$ patterns remaining.
+Each one may apply to any one of the $4$ [[Definition:Figure of Categorical Syllogism|figures]]
+Thus there are no more than $48$ [[Definition:Valid Argument|valid]] patterns of [[Definition:Categorical Syllogism|categorical syllogism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjunction implies Disjunction}
+Tags: Conjunction
+
+\begin{theorem}
+:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
+:$\begin{array}{|ccc|c|ccc|} \hline
+(p & \land & q) & \implies & (p & \lor & q) \\
+\hline
+F & F & F & T & F & F & F \\
+F & F & T & T & F & T & T \\
+T & F & F & T & T & T & F \\
+T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Angle Bisector Vector}
+Tags: Euclidean Geometry, Vector Algebra, Angle Bisector Vector
+
+\begin{theorem}
+Let $\mathbf u$ and $\mathbf v$ be [[Definition:Vector (Euclidean Space)|vectors]] of non-zero [[Definition:Vector Length|length]].
+Let $\left\Vert{\mathbf u}\right\Vert$ and $\left\Vert{\mathbf v}\right\Vert$ be their respective [[Definition:Vector Length|lengths]].
+Then $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the [[Definition:Angle Bisector|angle bisector]] of $\mathbf u$ and $\mathbf v$.
+\end{theorem}
+
+\begin{proof}
+:[[File:Angular Bisector Vector Diagram.png|400px]]
+As shown above:
+:Let the [[Definition:Angle|angle]] between $\mathbf u$ and $\mathbf v$ be $\gamma$.
+:Let the angle between $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\alpha$.
+:Let the angle between $\mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\beta$.
+Note that $\left\Vert{ \mathbf u }\right\Vert \mathbf v$ is $\mathbf v$ multiplied by the length of $\mathbf u$.
+By [[Vector Times Magnitude Same Length As Magnitude Times Vector]] the vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length.
+So $\left\Vert{\mathbf u}\right\Vert \mathbf v$, $\left\Vert{\mathbf v}\right\Vert \mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ make an [[Definition:Isosceles Triangle|isosceles triangle]].
+Therefore:
+{{begin-eqn}}
+{{eqn | l = 2 \beta + \alpha
+      | r = 180^\circ
+}}
+{{eqn | l = \beta
+      | r = 90^\circ - \frac 1 2 \alpha
+}}
+{{eqn | l = 2 \beta
+      | r = 180^\circ - \alpha
+}}
+{{end-eqn}}
+But since $\mathbf v$ and $\left\Vert{ \mathbf u }\right\Vert \mathbf v$ are [[Definition:Parallel Vectors|parallel]], we also have:
+{{begin-eqn}}
+{{eqn | l = \delta
+      | r = \alpha
+      | c = [[Parallelism implies Equal Corresponding Angles]]
+}}
+{{eqn | l = \delta + \gamma
+      | r = 180^\circ
+}}
+{{eqn | l = \alpha + \gamma
+      | r = 180^\circ
+}}
+{{eqn | l = \gamma
+      | r = 180^\circ - \alpha
+}}
+{{end-eqn}}
+Thus $\gamma = 2 \beta$, and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length from [[Vector Times Magnitude Same Length As Magnitude Times Vector]].
+Thus $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the diagonal of a [[Definition:Rhombus|rhombus]].
+ 
+The result follows from [[Diagonals of Rhombus Bisect Angles]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\mathbf a = \left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert\mathbf u$.
+Then:
+{{begin-eqn}}
+{{eqn | l=\cos \angle \mathbf u, \mathbf a
+      | r=\frac {\mathbf u \cdot \mathbf a} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= from [[Cosine Formula for Dot Product]]
+}}
+{{eqn | r=\frac {\mathbf u \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a} \right\Vert}
+      | c=
+}}
+{{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf u }\right) } {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= from [[Properties of Dot Product]]
+}}
+{{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf u }\right\Vert^2} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= from [[Dot Product of Vector with Itself]]
+}}
+{{eqn | r=\frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}
+      | c=
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l=\cos \angle \mathbf a, \mathbf v
+      | r=\frac {\mathbf v \cdot \mathbf a} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= from [[Cosine Formula for Dot Product]]
+}}
+{{eqn | r=\frac {\mathbf v \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c=
+}}
+{{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf v \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= from [[Properties of Dot Product]]
+}}
+{{eqn | r=\frac {\left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert^2} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}
+      | c= [[Dot Product of Vector with Itself]]
+}}
+{{eqn | r=\frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}
+      | c=
+}}
+{{end-eqn}}
+Comparing the two expressions gives us:
+:$\cos \angle \mathbf u, \mathbf a = \cos \angle \mathbf a, \mathbf v$
+Since the angle used in the dot product is always taken to be between $0$ and $\pi$ and [[Definition:Cosine|cosine]] is [[Definition:Injection|injective]] on this interval (from [[Shape of Cosine Function]]), we have:
+:$\angle \mathbf u, \mathbf a = \angle \mathbf a, \mathbf v$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Order 1 is Trivial}
+Tags: Subgroups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Then $\struct {G, \circ}$ has exactly $1$ [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $1$: the [[Definition:Trivial Subgroup|trivial subgroup]].
+\end{theorem}
+
+\begin{proof}
+From [[Trivial Subgroup is Subgroup]], $\struct {\set e, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$.
+Suppose $\struct {\set g, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$.
+From [[Group is not Empty]], $e \in \set g$.
+Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$.
+That is, $\struct {\set e, \circ}$ is the only [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ of [[Definition:Order of Structure|order]] $1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Injection from Finite Set to Itself is Surjection/Corollary}
+Tags: Injections, Permutation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $f: S \to S$ be an [[Definition:Injection|injection]].
+Then $f$ is a [[Definition:Permutation|permutation]].
+\end{theorem}
+
+\begin{proof}
+From [[Injection from Finite Set to Itself is Surjection]], $f$ is a [[Definition:Surjection|surjection]].
+As $f$ is thus both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], $f$ is a [[Definition:Bijection|bijection]] by definition.
+Thus as $f$ is a [[Definition:Bijection|bijection]] to itself, it is by definition a [[Definition:Permutation|permutation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Surjection from Finite Set to Itself is Permutation}
+Tags: Surjections, Permutation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $f: S \to S$ be an [[Definition:Surjection|surjection]].
+Then $f$ is a [[Definition:Permutation|permutation]].
+\end{theorem}
+
+\begin{proof}
+From [[Surjection iff Right Inverse]], $f$ has a [[Definition:Right Inverse Mapping|right inverse]] $g: S \to S$.
+From [[Right Inverse Mapping is Injection]], $g$ is an [[Definition:Injection|injection]].
+From [[Injection from Finite Set to Itself is Permutation]], $g$ is a [[Definition:Permutation|permutation]] and so a [[Definition:Bijection|bijection]].
+From [[Inverse of Bijection is Bijection]], $f$ is also a [[Definition:Bijection|bijection]].
+Thus as $f$ is a [[Definition:Bijection|bijection]] to itself, it is by definition a [[Definition:Permutation|permutation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Power of Group Element}
+Tags: Order of Group Elements
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$ be an [[Definition:Element|element]] of $G$ such that:
+:$\order g = n$
+where $\order g$ denotes the [[Definition:Order of Group Element|order]] of $g$.
+Then:
+:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n} }$
+where $\gcd \set {m, n}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+Let $\gcd \set {m, n} = d$.
+From [[Integers Divided by GCD are Coprime]]: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {g^m}^{n'}
+      | r = \paren {g^{d m'} }^{n'}
+      | c = Definition of $m'$
+}}
+{{eqn | r = \paren {g^{d n'} }^{m'}
+      | c = 
+}}
+{{eqn | r = \paren {g^n}^{m'}
+      | c = Definition of $n'$
+}}
+{{eqn | r = e^{m'}
+      | c = $n$ is the [[Definition:Order of Group Element|order]] of $g$
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+By [[Element to Power of Multiple of Order is Identity]]:
+:$\order {g^m} \divides n'$.
+{{AimForCont}} $\order {g^m} = n'' < n'$.
+By [[Bézout's Lemma]]:
+:$\exists x, y \in \Z: m x + n y = d$
+{{begin-eqn}}
+{{eqn | l = g^{d n''}
+      | r = g^{\paren {m x + n y} n''}
+      | c = 
+}}
+{{eqn | r = \paren {g^{m n''} }^x \paren {g^n}^{y n''}
+      | c = 
+}}
+{{eqn | r = e^x \paren {g^n}^{y n''}
+      | c = $n''$ is the [[Definition:Order of Group Element|order]] of $g^m$
+}}
+{{eqn | r = e^x e^{y n''}
+      | c = $n$ is the [[Definition:Order of Group Element|order]] of $g$
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+But $d n'' < d n' = n$, [[Definition:Contradiction|contradicting]] the fact that $n$ is the [[Definition:Order of Group Element|order]] of $g$.
+Therefore:
+:$\order {g^m} = n'$
+Recalling the definition of $n'$:
+:$\order {g^m} = \dfrac n {\gcd \set {m, n} }$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Group of Finite Order}
+Tags: Order of Groups
+
+\begin{theorem}
+Let $n \in \Z_{>0}$.
+Then there exists at least one [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is $n$.
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Cyclic Group of Order n]], there exists a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Order of Structure|order]] is $n$.
+In particular, a concrete example of such a [[Definition:Group|group]] is demonstrated in [[Roots of Unity under Multiplication form Cyclic Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vitali Theorem}
+Tags: Measure Theory
+
+\begin{theorem}
+There exists a [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] which is not [[Definition:Lebesgue Measure|Lebesgue measurable]].
+\end{theorem}
+
+\begin{proof}
+Let $\map \mu X$ denote the [[Definition:Lebesgue Measure|Lebesgue measure]] of a [[Definition:Set|set]] $X$ of [[Definition:Real Number|real numbers]].
+We have that:
+:$(1): \quad$ [[Lebesgue Measure is Countably Additive|$\map \mu X$ is a countably additive function]]
+:$(2): \quad$ [[Lebesgue Measure is Translation-Invariant|$\map \mu X$ is translation-invariant]]
+:$(3): \quad$ From [[Measure of Interval is Length]], $\map \mu {\closedint a b} = b - a$ for every [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+For all [[Definition:Real Number|real numbers]] in the [[Definition:Closed Unit Interval|closed unit interval]] $\mathbb I = \closedint 0 1$, define the [[Definition:Relation|relation]] $\sim$ such that:
+:$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$
+where $\Q$ is the set of [[Definition:Rational Number|rational numbers]].
+That is, $x \sim y$ {{iff}} their [[Definition:Real Subtraction|difference]] is [[Definition:Rational Number|rational]].
+By [[Non-Measurable Set of Real Numbers/Lemma 1|Non-Measurable Set of Real Numbers: Lemma 1]], $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+For each $x \in \mathbb I$, let $\eqclass x \sim$ denote the [[Definition:Equivalence Class|equivalence class]] of $x$ under $\sim$.
+By the [[Axiom:Axiom of Choice|Axiom of Choice]], it is possible to choose $1$ [[Definition:Element|element]] from each [[Definition:Equivalence Class|equivalence class]].
+Thus is created a [[Definition:Set|set]] $M \subset \mathbb I$ such that:
+:for each $x \in R$ there exists a [[Definition:Unique|unique]] $y \in M$ and $r \in \Q$ such that $x = y + r$.
+Let:
+:$M_r = \set {y + r: y \in M}$
+for each $r \in \Q$.
+Thus $\R$ is [[Definition:Partition (Set Theory)|partitioned]] into [[Definition:Countable|countably many]] [[Definition:Disjoint Sets|disjoint sets]]:
+:$(1): \quad \displaystyle \R = \bigcup \set {M_r: r \in \Q}$
+Suppose $M$ were [[Definition:Lebesgue Measure|Lebesgue measurable]].
+First, $\map \mu M = 0$ is impossible, because from $(1)$ this would mean  $\map \mu \R = 0$.
+Suppose $\map \mu M > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \mu {\closedint 0 1}
+      | o = \ge
+      | r = \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }
+      | c = 
+}}
+{{eqn | r = \sum_{r \mathop \in \Q \mathop \land 0 \mathop \le r \mathop \le 1} \map \mu {M_r}
+      | c = 
+}}
+{{eqn | r = \infty
+      | c = as each $M_r$ would have to have the same measure as $M$
+}}
+{{end-eqn}}
+So $\map \mu M > 0$ is also impossible.
+Hence $M$ is not [[Definition:Lebesgue Measure|Lebesgue measurable]].
+{{qed}}
+{{refactor|Put this statement into a "example" page or something}}
+The set $M$ described here is an example of a [[Definition:Vitali Set|Vitali set]]: a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] which has no [[Definition:Lebesgue Measure|Lebesgue measure]].
+{{AoC}}
+{{Namedfor|Giuseppe Vitali|cat = Vitali}}
+\end{proof}<|endoftext|>
+\section{Finite Number of Groups of Given Finite Order}
+Tags: Order of Groups
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then there exists a [[Definition:Finite|finite number]] of [[Definition:Group Type|types of group]] of [[Definition:Order of Structure|order $n$]].
+\end{theorem}
+
+\begin{proof}
+For any [[Definition:Group|group]] $\struct {G, \circ}$ of [[Definition:Order of Structure|order $n$]] and any [[Definition:Set|set]] of $n$ [[Definition:Element|elements]], $X$ can be the [[Definition:Underlying Set|underlying set]] of a [[Definition:Group|group]] which is [[Definition:Group Isomorphism|isomorphic]] to $\struct {G, \circ}$, as follows:
+Choose a [[Definition:Bijection|bijection]] $\phi: G \to X$.
+Define the [[Definition:Group Operation|group operation]] $*$ on $X$ by the rule:
+:$\map \phi {g_1} * \map \phi {g_2} = \map \phi {g_1 \circ g_2}$
+for all $g_1, g_2 \in G$.
+By definition, $\phi$ is an [[Definition:Group Isomorphism|isomorphism]].
+From [[Isomorphism Preserves Groups]] it follows that $\struct {X, *}$ is a [[Definition:Group|group]].
+Thus all [[Definition:Group|groups]] of [[Definition:Order of Structure|order $n$]] of all possible [[Definition:Group Type|types]] appear among all possible arrangements of [[Definition:Binary Operation|binary operation]].
+But from [[Count of Binary Operations on Set]], this count is $n^{\paren {n^2} }$.
+This is the [[Definition:Upper Bound of Set|upper bound]] to the number of [[Definition:Group Type|types of group]] of [[Definition:Order of Structure|order $n$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nu of Prime Number is 1}
+Tags: Prime Groups, Nu Function
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Then:
+:$\map \nu p = 1$
+where $\nu$ denotes the [[Definition:Nu Function|$\nu$ function]]: the number of [[Definition:Group Type|types of group]] of a given [[Definition:Order of Group|order]].
+\end{theorem}
+
+\begin{proof}
+Let $G_1$ and $G_2$ be [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $p$.
+From [[Prime Group is Cyclic]], $G_1$ and $G_2$ are both [[Definition:Cyclic Group|cyclic groups]].
+From [[Cyclic Groups of Same Order are Isomorphic]], $G_1$ and $G_2$ are [[Definition:Group Isomorphism|isomorphic]].
+Thus by definition, $G_1$ and $G_2$ are of the same [[Definition:Group Type|type]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Subset under Relation equals Union of Images of Elements}
+Tags: Relation Theory, Subsets, Set Union
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
+Let $X \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$\displaystyle \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$
+where:
+:$\RR \sqbrk X$ is the [[Definition:Image of Subset under Relation|image of the subset $X$ under $\RR$]]
+:$\map \RR x$ is the [[Definition:Image of Element under Relation|image of the element $x$ under $\RR$]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$
+:$\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$
+First:
+{{begin-eqn}}
+{{eqn | l = y 
+      | o = \in
+      | r = \RR \sqbrk X
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x \in X: \tuple {x, y}
+      | o = \in
+      | r = \RR
+      | c = Definition of $\RR \sqbrk X$
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x, y}
+      | o = \in
+      | r = \bigcup_{x \mathop \in X} \set {\tuple {x, y} \in \RR}
+      | c = {{Defof|Set Union}}
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \bigcup_{x \mathop \in X} \set {y \in T: \tuple {x, y} \in \RR}
+      | c = {{Defof|Relation}}
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \bigcup_{x \mathop \in X} \map \RR x
+      | c = Definition of $\map \RR x$
+}}
+{{eqn | ll= \leadsto
+      | l = \RR \sqbrk X
+      | o = \subseteq
+      | r = \bigcup_{x \mathop \in X} \map \RR x
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \bigcup_{x \mathop \in X} \map \RR x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \bigcup_{x \mathop \in X} \set {y \in T: \tuple {x, y} \in \RR}
+      | c = Definition of $\map \RR x$
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x \in X: \tuple {x, y}
+      | o = \in
+      | r = \RR
+      | c = {{Defof|Set Union}}
+}}
+{{eqn | ll= \leadsto
+      | l = y 
+      | o = \in
+      | r = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}
+      | c = {{Defof|Relation}}
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \RR \sqbrk X
+      | c = Definition of $\RR \sqbrk X$
+}}
+{{eqn | ll= \leadsto
+      | l = \bigcup_{x \mathop \in X} \map \RR x
+      | o = \subseteq
+      | r = \RR \sqbrk X
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \bigcup_{x \mathop \in X} \map \RR x \subseteq \RR \sqbrk X$
+and:
+:$\displaystyle \RR \sqbrk X \subseteq \bigcup_{x \mathop \in X} \map \RR x$
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+[[Category:Relation Theory]]
+[[Category:Subsets]]
+[[Category:Set Union]]
+r4xlio6zh4r18w9i8hn8mtcy1wot3bh
+\end{proof}<|endoftext|>
+\section{Preimage of Subset under Relation equals Union of Preimages of Elements}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
+Let $\RR^{-1} \subseteq T \times S$ be the [[Definition:Inverse Relation|inverse relation]] to $\RR$
+Let $Y \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
+Then:
+:$\RR^{-1} \sqbrk Y = \displaystyle \bigcup_{y \mathop \in Y} \map {\RR^{-1} } y$
+where:
+:$\RR^{-1} \sqbrk Y$ is the [[Definition:Preimage of Subset under Relation|preimage]] of the [[Definition:Subset|subset]] $Y$ under $\RR$
+:$\RR^{-1} \sqbrk Y$ is the [[Definition:Preimage of Element under Relation|preimage]] of the [[Definition:Element|element]] $y$ under $\RR$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\RR^{-1} \subseteq T \times S$ is a [[Definition:Relation|relation]] on $T \times S$.
+Thus [[Image of Subset under Relation equals Union of Images of Elements]] can be applied directly.
+{{qed}}
+[[Category:Relation Theory]]
+qe7ehp1ro09ic0fjgyeilan3myg8mvl
+\end{proof}<|endoftext|>
+\section{Preimage of Subset under Mapping equals Union of Preimages of Elements}
+Tags: Preimages under Mappings, Subsets, Set Union
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]] from $S$ to $T$.
+Let $f^{-1} \subseteq T \times S$ be the [[Definition:Inverse of Mapping|inverse]] of $f$, defined as:
+:$f^{-1} = \set {\tuple {t, s}: \map f s = t}$
+Let $Y \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
+Then:
+:$\displaystyle f^{-1} \sqbrk Y = \bigcup_{y \mathop \in Y} \map {f^{-1} } y$
+where:
+:$f^{-1} \sqbrk Y$ is the [[Definition:Preimage of Subset under Mapping|preimage of the subset $Y$ under $f$]]
+:$\map {f^{-1} } y$ is the [[Definition:Preimage of Element under Mapping|preimage of the element $y$ under $f$]].
+\end{theorem}
+
+\begin{proof}
+By definition, $f^{-1} \subseteq T \times S$ is a [[Definition:Relation|relation]] on $T \times S$.
+Thus [[Image of Subset under Relation equals Union of Images of Elements]] can be applied directly.
+{{qed}}
+[[Category:Preimages under Mappings]]
+[[Category:Subsets]]
+[[Category:Set Union]]
+a3ik6tclkfa5owlyn526l44dveapcw9
+\end{proof}<|endoftext|>
+\section{Image of Domain of Mapping is Image Set}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+The [[Definition:Image of Subset under Mapping|image of $S$]] is the [[Definition:Image of Mapping|image set of $f$]]:
+:$f \sqbrk S = \Img f$
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]].
+Thus [[Image of Domain of Relation is Image Set]] applies.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Singleton under Mapping}
+Tags: Mapping Theory, Singletons
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then the [[Definition:Image of Element under Mapping|image]] of an [[Definition:Element|element]] of $S$ is equal to the [[Definition:Image of Subset under Mapping|image]] of a [[Definition:Singleton|singleton]] containing that element, the singleton being a [[Definition:Subset|subset]] of $S$:
+:$\forall s \in S: \set {\map f s} = f \sqbrk {\set s}$
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]].
+Thus [[Image of Singleton under Relation]] applies.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Image of Subset under Mapping equals Union of Images of Elements}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]] from $S$ to $T$.
+Let $X \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$\displaystyle f \sqbrk X = \bigcup_{x \mathop \in X} \map f x$
+where:
+:$f \sqbrk X$ is the [[Definition:Image of Subset under Mapping|image of the subset $X$ under $f$]]
+:$\map f x$ is the [[Definition:Image of Element under Mapping|image of the element $x$ under $f$]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]].
+Thus [[Image of Subset under Relation equals Union of Images of Elements]] applies.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Mapping Images are Disjoint only if Domains are Disjoint}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let:
+:$f \sqbrk S \cap f \sqbrk T = \O$
+where $f \sqbrk S$ denotes the [[Definition:Image Set of Mapping|image set]] of $S$.
+Then:
+:$S \cap T = \O$
+\end{theorem}
+
+\begin{proof}
+From [[Image of Intersection under Mapping]]:
+:$f \sqbrk {S \cap T} \subseteq f \sqbrk S \cap f \sqbrk T$
+From [[Empty Set is Subset of All Sets]]:
+:$f \sqbrk {S \cap T} = \O$
+From [[Image of Subset under Relation is Subset of Image/Corollary 2|Image of Subset under Relation is Subset of Image: Corollary 2]]:
+:$S \cap T = \O$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Relation is Domain of Inverse Relation}
+Tags: Relation Theory, Inverse Relations
+
+\begin{theorem}
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
+Let $\RR^{-1} \subseteq T \times S$ be the [[Definition:Inverse Relation|inverse of $\RR$]].
+Then:
+:$\Img \RR = \Dom {\RR^{-1} }$
+That is, the [[Definition:Image of Relation|image]] of a [[Definition:Relation|relation]] is the [[Definition:Domain of Relation|domain]] of its [[Definition:Inverse Relation|inverse]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\Img \RR := \set {t \in T: \exists s \in S: \tuple {s, t} \in \RR}$
+:$\Dom {\RR^{-1} } := \set {t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1} }$
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \Img \RR
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists s \in S: \tuple {s, x}
+      | o = \in
+      | r = \RR
+      | c = {{Defof|Image of Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists s \in S: \tuple {x, s}
+      | o = \in
+      | r = \RR^{-1}
+      | c = {{Defof|Inverse Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \Dom {\RR^{-1} }
+      | c = {{Defof|Domain of Relation}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Relation is Image of Inverse Relation}
+Tags: Relation Theory, Inverse Relations
+
+\begin{theorem}
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
+Let $\RR^{-1} \subseteq T \times S$ be the [[Definition:Inverse Relation|inverse of $\RR$]].
+Then:
+:$\Dom \RR = \Img {\RR^{-1} }$
+That is, the [[Definition:Domain of Relation|domain]] of a [[Definition:Relation|relation]] is the [[Definition:Image of Relation|image]] of its [[Definition:Inverse Relation|inverse]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\Dom \RR := \set {s \in S: \exists t \in T: \tuple {s, t} \in \RR}$
+:$\Img {\RR^{-1} } := \set {s \in S: \exists T \in T: \tuple {t, s} \in \RR^{-1} }$
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \Dom \RR
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists t \in T: \tuple {x, t}
+      | o = \in
+      | r = \RR
+      | c = {{Defof|Domain of Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists t \in T: \tuple {t, x}
+      | o = \in
+      | r = \RR^{-1}
+      | c = {{Defof|Inverse Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \Img {\RR^{-1} }
+      | c = {{Defof|Image of Relation}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element in Preimage of Image under Mapping}
+Tags: Preimages under Mappings
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+:$\forall x \in S: x \in f^{-1} \sqbrk {\map f x}$
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Mapping|mapping]] is by definition a [[Definition:Left-Total Relation|left-total relation]].
+Therefore [[Preimage of Image under Left-Total Relation is Superset]] applies:
+:$A \subseteq S \implies A \subseteq f^{-1} \sqbrk {f \sqbrk A}$
+Thus:
+:$\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element in Image of Preimage under Mapping}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+:$\forall y \in T: \in f \sqbrk {f^{-1} \sqbrk y} = \set y$
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Mapping|mapping]] is by definition a [[Definition:Relation|relation]].
+Therefore [[Image of Preimage under Mapping/Corollary|Image of Preimage under Mapping: Corollary]] applies:
+:$B \subseteq \Img S \implies \paren {f \circ f^{-1} } \sqbrk B = B$
+Thus:
+:$\set y \subseteq T \implies f^{-1} \sqbrk {f \sqbrk {\set y} } = \set y$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Set Difference under Injection}
+Tags: Set Difference, Injections
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$.
+Let $S_1 \setminus S_2$ denote the [[Definition:Set Difference|set difference]] between $S_1$ and $S_2$.
+Then:
+:$\forall S_1, S_2 \subseteq S: f \left[{S_1}\right] \setminus f \left[{S_2}\right] = f \left[{S_1 \setminus S_2}\right]$
+{{iff}} $f$ is an [[Definition:Injection|injection]].
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Injection|injection]] is a type of [[Definition:One-to-One Relation|one-to-one relation]], and therefore also a [[Definition:One-to-Many Relation|one-to-many relation]].
+Therefore [[One-to-Many Image of Set Difference]] applies:
+:$\mathcal R \left[{S_1}\right] \setminus \mathcal R \left[{S_2}\right] = \mathcal R \left[{S_1 \setminus S_2}\right]$
+{{iff}} $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]].
+We have that $f$ is a [[Definition:Mapping|mapping]] and therefore a [[Definition:Many-to-One Relation|many-to-one relation]].
+So $f$ is a [[Definition:One-to-Many Relation|one-to-many relation]] {{iff}} $f$ is also an [[Definition:Injection|injection]].
+It follows that:
+:$\forall S_1, S_2 \subseteq S: f \left[{S_1}\right] \setminus f \left[{S_2}\right] = f \left[{S_1 \setminus S_2}\right]$
+{{iff}} $f$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Union}
+Tags: De Morgan's Laws, Indexed Families
+
+\begin{theorem}
+:$\displaystyle \map \complement {\bigcup_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \map \complement {S_i}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \complement {\bigcup_{i \mathop \in I} S_i}
+      | r = \mathbb U \setminus \paren {\bigcup_{i \mathop \in I} S_i}
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | r = \bigcap_{i \mathop \in I} \paren {\mathbb U \setminus S_i}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union|De Morgan's Laws for Set Difference: Difference with Union]]
+}}
+{{eqn | r = \bigcap_{i \mathop \in I} \map \complement {S_i}
+      | c = {{Defof|Set Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection}
+Tags: De Morgan's Laws, Indexed Families
+
+\begin{theorem}
+:$\displaystyle \map \complement {\bigcap_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \map \complement {S_i}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \complement {\bigcap_{i \mathop \in I} S_i}
+      | r = \mathbb U \setminus \paren {\bigcap_{i \mathop \in I} S_i}
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \paren {\mathbb U \setminus S_i}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's Laws for Set Difference: Difference with Intersection]]
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \map \complement {S_i}
+      | c = {{Defof|Set Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Sets Exists iff Limit Inferior contains Limit Superior}
+Tags: Limits Superior of Set Sequences, Limits Inferior of Set Sequences
+
+\begin{theorem}
+Let $\Bbb S = \set {E_n : n \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]].
+Then $\Bbb S$ [[Definition:Limit of Sets|converges to a limit]] {{iff}}:
+:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\Bbb S$ [[Definition:Limit of Sets|converge to a limit]].
+Then by definition:
+:$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$
+and so by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that:
+:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
+From [[Limit Superior includes Limit Inferior]]:
+:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$
+whether or not $\Bbb S$ [[Definition:Limit of Sets|converges to a limit]].
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$
+and so by definition $\Bbb S$ [[Definition:Limit of Sets|converges to a limit]].
+{{qed}}
+[[Category:Limits Superior of Set Sequences]]
+[[Category:Limits Inferior of Set Sequences]]
+is5qz1szkina5763lfyl7zdoq1ymgtz
+\end{proof}<|endoftext|>
+\section{Closure of Intersection and Symmetric Difference imply Closure of Set Difference}
+Tags: Set Intersection, Set Difference, Symmetric Difference, Set Systems
+
+\begin{theorem}
+Let $\mathcal R$ be a [[Definition:System of Sets|system of sets]] such that for all $A, B \in \mathcal R$:
+: $(1): \quad A \cap B \in \mathcal R$
+: $(2): \quad A * B \in \mathcal R$
+where $\cap$ denotes [[Definition:Set Intersection|set intersection]] and $*$ denotes [[Definition:Symmetric Difference|set symmetric difference]].
+Then:
+:$\forall A, B \in \mathcal R: A \setminus B \in \mathcal R$
+where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \mathcal R$.
+From [[Set Difference as Symmetric Difference with Intersection]]:
+: $A * \left({A \cap B}\right) = A \setminus B$
+By hypothesis:
+:$A \cap B \in \mathcal R$
+and:
+:$A * \left({A \cap B}\right) \in \mathcal R$
+and so:
+:$A \setminus B \in \mathcal R$
+{{qed}}
+[[Category:Set Intersection]]
+[[Category:Set Difference]]
+[[Category:Symmetric Difference]]
+[[Category:Set Systems]]
+cgvyaoug8dsj4nteepq16t572lqwan0
+\end{proof}<|endoftext|>
+\section{Closure of Intersection and Symmetric Difference imply Closure of Union}
+Tags: Set Intersection, Set Union, Symmetric Difference, Set Systems
+
+\begin{theorem}
+Let $\mathcal R$ be a [[Definition:System of Sets|system of sets]] such that for all $A, B \in \mathcal R$:
+: $(1): \quad A \cap B \in \mathcal R$
+: $(2): \quad A * B \in \mathcal R$
+where $\cap$ denotes [[Definition:Set Intersection|set intersection]] and $*$ denotes [[Definition:Symmetric Difference|set symmetric difference]].
+Then:
+:$\forall A, B \in \mathcal R: A \cup B \in \mathcal R$
+where $\cup$ denotes [[Definition:Set Union|set union]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \mathcal R$.
+From [[Union as Symmetric Difference with Intersection]]:
+: $\left({A * B}\right) * \left({A \cap B}\right) = A \cup B$
+By hypothesis:
+:$A \cap B \in \mathcal R$
+and:
+:$\left({A * B}\right) * \left({A \cap B}\right)$
+and so:
+:$A \cup B \in \mathcal R$
+{{qed}}
+[[Category:Set Intersection]]
+[[Category:Set Union]]
+[[Category:Symmetric Difference]]
+[[Category:Set Systems]]
+lztpo7atayaw90asuztd0i0luzejh0o
+\end{proof}<|endoftext|>
+\section{Empty Set and Set form Algebra of Sets}
+Tags: Algebras of Sets
+
+\begin{theorem}
+Let $S$ be any [[Definition:Non-Empty Set|non-empty set]].
+Then $\left\{{S, \varnothing}\right\}$ is (trivially) an [[Definition:Algebra of Sets|algebra of sets]], where $S$ is the [[Definition:Unit of System of Sets|unit]].
+\end{theorem}
+
+\begin{proof}
+From [[Union is Idempotent]]:
+:$S \cup S = S$
+and
+:$\varnothing \cup \varnothing = \varnothing$
+Then from [[Union with Empty Set]]:
+:$S \cup \varnothing = S$
+So $\left\{{S, \varnothing}\right\}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Set Union|union]].
+From [[Relative Complement of Empty Set]]:
+:$\complement_S \left({\varnothing}\right) = S$
+and from [[Relative Complement with Self is Empty Set]]:
+:$\complement_S \left({S}\right) = \varnothing$
+so $\left\{{S, \varnothing}\right\}$ is [[Definition:Closed under Mapping|closed]] under [[Definition:Set Complement|complement]].
+Hence the result, by definition of [[Definition:Algebra of Sets/Definition 1|algebra of sets]].
+{{qed}}
+[[Category:Algebras of Sets]]
+13vhg5iqcdka84nj8odh2tinqsasb7e
+\end{proof}<|endoftext|>
+\section{Closure of Union and Complement imply Closure of Set Difference}
+Tags: Set Union, Set Difference, Set Complement, Set Systems
+
+\begin{theorem}
+Let $\RR$ be a [[Definition:System of Sets|system of sets]] on a [[Definition:Universal Set|universe]] $\mathbb U$ such that for all $A, B \in \RR$:
+:$(1): \quad A \cup B \in \RR$
+:$(2): \quad \map \complement A \in \RR$
+where $\cup$ denotes [[Definition:Set Union|set union]] and $\complement$ denotes [[Definition:Set Complement|complement]] (relative to $\mathbb U$).
+Then:
+:$\forall A, B \in \RR: A \setminus B \in \RR$
+where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \RR$.
+{{begin-eqn}}
+{{eqn | l = A \setminus B
+      | r = A \cap \map \complement B
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{eqn | r = \map \complement {\map \complement A \cup B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+As both [[Definition:Set Union|set union]] and [[Definition:Set Complement|complement]] are [[Definition:Closed Operation|closed]] in $\RR$ the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Limit Inferior is Limit Superior of Complements}
+Tags: Limits Superior of Set Sequences, Limits Inferior of Set Sequences
+
+\begin{theorem}
+Let $\left\{{E_n : n \in \N}\right\}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]].
+Then:
+:$\displaystyle \complement \left({\liminf_{n \mathop \to \infty} \ E_n}\right) = \limsup_{n \mathop \to \infty} \ \complement \left({E_n}\right)$
+where $\liminf$ and $\limsup$ denote the [[Definition:Limit Inferior of Sequence of Sets|limit inferior]] and [[Definition:Limit Superior of Sequence of Sets|limit superior]], respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \complement \left({\liminf_{n \mathop \to \infty} \ E_n}\right)
+      | r = \complement \left({\bigcup_{n \mathop = 0}^\infty \bigcap_{i \mathop = n}^\infty E_n}\right)
+      | c = {{Defof|Limit Inferior of Sequence of Sets|index = 1}}
+}}
+{{eqn | r = \bigcap_{n \mathop = 0}^\infty \complement \left({\bigcap_{i \mathop = n}^\infty E_n}\right)
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Union|De Morgan's Laws]]
+}}
+{{eqn | r = \bigcap_{n \mathop = 0}^\infty \bigcup_{i \mathop = n}^\infty \complement \left({E_n}\right)
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection|De Morgan's Laws]]
+}}
+{{eqn | r = \limsup_{n \mathop \to \infty} \ \complement \left({E_n}\right)
+      | c = {{Defof|Limit Superior of Sequence of Sets|index = 1}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Limit Superior is Limit Inferior of Complements}
+Tags: Limits Superior of Set Sequences, Limits Inferior of Set Sequences
+
+\begin{theorem}
+Let $\left\{{E_n : n \in \N}\right\}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]].
+Then:
+:$\displaystyle \complement \left({\limsup_{n \to \infty} \ E_n}\right) = \liminf_{n \to \infty} \ \complement \left({E_n}\right)$
+where $\limsup$ and $\liminf$ denote the [[Definition:Limit Superior of Sequence of Sets|limit superior]] and [[Definition:Limit Inferior of Sequence of Sets|limit inferior]], respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn|l = \complement \left({\limsup_{n \to \infty} \ E_n}\right)
+     |r = \complement \left({\bigcap_{n \mathop = 0}^\infty \bigcup_{i \mathop = n}^\infty E_n}\right)
+     |c = Definition of [[Definition:Limit Superior of Sequence of Sets/Definition 1|limit superior]]
+}}
+{{eqn|r = \bigcup_{n \mathop = 0}^\infty \complement \left({\bigcup_{i \mathop = n}^\infty E_n}\right)
+     |c = [[De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection|De Morgan's Laws]]
+}}
+{{eqn|r = \bigcup_{n \mathop = 0}^\infty \bigcap_{i \mathop = n}^\infty \complement \left({E_n}\right)
+     |c = [[De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Union|De Morgan's Laws]]
+}}
+{{eqn|r = \liminf_{n \to \infty} \ \complement \left({E_n}\right)
+     |c = Definition of [[Definition:Limit Inferior of Sequence of Sets/Definition 1|limit inferior]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sigma-Ring is Closed under Countable Intersections}
+Tags: Sigma-Rings
+
+\begin{theorem}
+Let $\mathcal R$ be a [[Definition:Sigma-Ring|$\sigma$-ring]].
+Let $\left \langle{A_n}\right \rangle_{n \mathop \in \N} \in \mathcal R$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]] in $\mathcal R$.
+Then:
+:$\displaystyle \bigcap_{n \mathop = 1}^\infty A_n \in \mathcal R$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \N_{>0}
+      | l = A_1, A_n \in \mathcal R
+      | o = \implies
+      | r = A_1 \setminus A_n \in \mathcal R
+      | c = Axiom $(SR2)$ for [[Definition:Sigma-Ring|$\sigma$-rings]]
+}}
+{{eqn | o = \implies
+      | r = \bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) \in \mathcal R
+      | c = Axiom $(SR3)$ for [[Definition:Sigma-Ring|$\sigma$-rings]]
+}}
+{{eqn | o = \implies
+      | r = A_1 \setminus \left({\bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) }\right) \in \mathcal R
+      | c = Axiom $(SR2)$ for [[Definition:Sigma-Ring|$\sigma$-rings]]
+}}
+{{end-eqn}}
+From [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's laws: Difference with Intersection]]:
+:$\displaystyle \bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) = A_1 \setminus \left({\bigcap_{n \mathop = 2}^\infty A_n}\right)$
+From [[Set Difference with Set Difference]]:
+{{begin-eqn}}
+{{eqn | l = A_1 \setminus \left({A_1 \setminus \left({\displaystyle \bigcap_{n \mathop = 2}^\infty A_n}\right) }\right)
+      | r = A_1 \cap \left({\displaystyle \bigcap_{n \mathop = 2}^\infty A_n}\right)
+}}
+{{eqn | r = {\bigcap_{n \mathop = 1}^\infty A_n}
+}}
+{{end-eqn}}
+Combining the previous equalities, it follows that:
+: $\displaystyle \bigcap_{n \mathop = 1}^\infty A_n \in \mathcal R$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pappus's Hexagon Theorem}
+Tags: Euclidean Geometry, Projective Geometry
+
+\begin{theorem}
+Let $A, B, C$ be a [[Definition:Set|set]] of [[Definition:Collinear Points|collinear points]].
+Let $a, b, c$ be another [[Definition:Set|set]] of [[Definition:Collinear Points|collinear points]].
+Let $X, Y, Z$ be the [[Definition:Intersection (Geometry)|points of intersection]] of each of the [[Definition:Straight Line|straight lines]] $Ab$ and $aB$, $Ac$ and $aC$, and $Bc$ and $bC$.
+Then $X, Y, Z$ are [[Definition:Collinear Points|collinear points]].
+\end{theorem}
+
+\begin{proof}
+:[[File:PappusHexagonTheorem.png|400px]]
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Plane contains Infinite Number of Lines}
+Tags: Projective Geometry
+
+\begin{theorem}
+A [[Definition:Plane|plane]] contains an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Straight Line|lines]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Plane|plane]] contains an [[Definition:Infinite Set|infinite number]] of [[Definition:Point|points]].
+Not all these [[Definition:Point|points]] are [[Definition:Collinear|collinear]].
+Let $A$, $B$ and $C$ be [[Definition:Point|points]] in a [[Definition:Plane|plane]] $P$.
+From [[Axiom:Propositions of Incidence/Line in Plane|Propositions of Incidence: Line in Plane]], any two of these [[Definition:Point|points]] determine a [[Definition:Straight Line|line]].
+Consider the [[Definition:Straight Line|lines]] $AB$, $AC$ and $BC$, all of which are [[Definition:Distinct Elements|distinct]].
+Let $X$ be one of the [[Definition:Infinite Set|infinite number]] of [[Definition:Point|points]] on $BC$ which is not $B$ or $C$.
+Then $AX$ is a [[Definition:Straight Line|line]] in $P$ which is [[Definition:Distinct Elements|distinct]] from both $AB$ and $AC$.
+As there is are [[Definition:Infinite Set|infinite number]] of [[Definition:Point|points]] on $BC$, there are an [[Definition:Infinite Set|infinite number]] of [[Definition:Straight Line|lines]] incident to $A$ and $BC$.
+All these [[Definition:Straight Line|lines]] are in $P$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two Planes have Line in Common}
+Tags: Projective Geometry
+
+\begin{theorem}
+Two [[Definition:Distinct Elements|distinct]] [[Definition:Plane|planes]] have exactly one [[Definition:Straight Line|(straight) line]] in common.
+\end{theorem}
+
+\begin{proof}
+Take two [[Definition:Distinct Elements|distinct]] [[Definition:Straight Line|lines]] in [[Definition:Plane|plane]] $1$.
+From [[Axiom:Propositions of Incidence/Plane and Line|Propositions of Incidence: Plane and Line]], they each meet [[Definition:Plane|plane]] $2$ in one point each, say at $A$ and $B$.
+Thus $A$ and $B$ both lie in both [[Definition:Plane|planes]].
+Thus the [[Definition:Straight Line|line]] defined by $A$ and $B$ lies in both [[Definition:Plane|planes]].
+{{qed}}
+{{Handwaving}}
+\end{proof}<|endoftext|>
+\section{Three Non-Collinear Planes have One Point in Common}
+Tags: Projective Geometry
+
+\begin{theorem}
+Three [[Definition:Plane|planes]] which are not [[Definition:Collinear Planes|collinear]] have exactly one [[Definition:Point|point]] in all three [[Definition:Plane|planes]].
+\end{theorem}
+
+\begin{proof}
+Let $A$, $B$ and $C$ be the three [[Definition:Plane|planes]] in question.
+From [[Two Planes have Line in Common]], $A$ and $B$ share a [[Definition:Straight Line|line]], $p$ say.
+From [[Axiom:Propositions of Incidence/Plane and Line|Propositions of Incidence: Plane and Line]], $p$ meets $C$ in one [[Definition:Point|point]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Duality in the Plane}
+Tags: Projective Geometry
+
+\begin{theorem}
+Let $P$ be a [[Definition:Theorem|theorem]] of [[Definition:Projective Geometry|projective geometry]] proven using the [[Axiom:Propositions of Incidence|propositions of incidence]].
+Let $Q$ be the [[Definition:Statement|statement]] created from $P$ by interchanging:
+:$(1): \quad$ the terms '''[[Definition:Point|point]]''' and '''[[Definition:Straight Line|(straight) line]]'''
+:$(2): \quad$ the terms '''[[Definition:Collinear Points|collinear]]''' (of [[Definition:Point|points]]) and '''[[Definition:Concurrent Lines|concurrent]]''' (of [[Definition:Straight Line|lines]])
+:$(3): \quad$ the terms '''lie on''' and '''[[Definition:Intersection (Geometry)|intersect]] at'''
+and so on.
+Then $Q$ is also a [[Definition:Theorem|theorem]] of [[Definition:Projective Geometry|projective geometry]].
+\end{theorem}<|endoftext|>
+\section{Principle of Duality in Space}
+Tags: Projective Geometry
+
+\begin{theorem}
+Let $P$ be a [[Definition:Theorem|theorem]] of [[Definition:Projective Geometry|projective geometry]] proven using the [[Axiom:Propositions of Incidence|propositions of incidence]].
+Let $Q$ be the [[Definition:Statement|statement]] created from $P$ by interchanging:
+:$(1) \quad$ the terms '''[[Definition:Point|point]]''' and '''[[Definition:Plane|plane]]'''
+:$(2) \quad$ the terms '''lie on''' and '''[[Definition:Intersection (Geometry)|intersect]] at'''
+and so on.
+Then $Q$ is also a [[Definition:Theorem|theorem]] of [[Definition:Projective Geometry|projective geometry]].
+\end{theorem}<|endoftext|>
+\section{Desargues' Theorem}
+Tags: Projective Geometry
+
+\begin{theorem}
+Let $\triangle ABC$ and $\triangle A'B'C'$ be [[Definition:Triangle (Geometry)|triangles]] lying in the same or different [[Definition:Plane|planes]].
+Let the [[Definition:Straight Line|lines]] $AA'$, $BB'$ and $CC'$ [[Definition:Intersection (Geometry)|intersect]] in the [[Definition:Point|point]] $O$.
+Then $BC$ meets $B'C'$ in $L$, $CA$ meets $C'A'$ in $M$ and $AB$ meets $A'B'$ in $N$, where $L, M, N$ are [[Definition:Collinear Points|collinear]].
+\end{theorem}
+
+\begin{proof}
+:[[File:DesarguesTheorem.png|500px]]
+Let $\triangle ABC$ and $\triangle A'B'C'$ be in different [[Definition:Plane|planes]] $\pi$ and $\pi'$ respectively.
+Since $BB'$ and $CC'$ [[Definition:Intersection (Geometry)|intersect]] in $O$, it follows that $B$, $B'$, $C$ and $C'$ lie in a [[Definition:Plane|plane]].
+Thus $BC$ must meet $B'C'$ in a [[Definition:Point|point]] $L$.
+By the same argument, $CA$ meets $C'A'$ in a [[Definition:Point|point]] $M$ and $AB$ meets $A'B'$ in a [[Definition:Point|point]] $N$.
+These [[Definition:Point|points]] $L, M, N$ are in each of the [[Definition:Plane|planes]] $\pi$ and $\pi'$.
+By [[Two Planes have Line in Common]] they are therefore [[Definition:Collinear Points|collinear]] on the line where $\pi$ and $\pi'$ meet.
+Now let $\triangle ABC$ and $\triangle A'B'C'$ be in the same [[Definition:Plane|plane]] $\pi$.
+Let $OPP'$ be any [[Definition:Straight Line|line]] through $O$ which does not lie in $\pi$.
+Then since $PP'$ meets $AA'$ in $O$, the four [[Definition:Point|points]] $P, P', A, A$ are [[Definition:Coplanar Points|coplanar]].
+Thus $PA$ meets $P'A'$ at a [[Definition:Point|point]] $A''$.
+Similarly $PB$ meets $P'B'$ at a [[Definition:Point|point]] $B''$, and $PC$ meets $P'C'$ at a [[Definition:Point|point]] $C''$.
+The lines $BC, B'C'$ and $B''C''$ are the three [[Definition:Straight Line|lines]] of [[Definition:Intersection (Geometry)|intersection]] of the three [[Definition:Plane|planes]] $PBC$, $P'B'C'$ and $\pi$ taken in pairs.
+So $BC$, $B'C'$ and $B''C''$ meet in a [[Definition:Point|point]] $L$.
+Similarly $CA$, $C'A'$ and $C''A''$ meet in a [[Definition:Point|point]] $M$ and $AB$, $A'B'$ and $A''B''$ meet in a [[Definition:Point|point]] $N$.
+The two [[Definition:Triangle (Geometry)|triangles]] $\triangle ABC$ and $\triangle A''B''C''$ are in different [[Definition:Plane|planes]], and $AA''$, $BB''$ and $CC''$ meet in $P$.
+Thus $L$, $M$ and $N$ are [[Definition:Collinear Points|collinear]] by the first part of this proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Mappings which Agree is Mapping}
+Tags: Mapping Theory, Set Union
+
+\begin{theorem}
+Let $A, B, Y$ be [[Definition:Set|sets]].
+Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Mapping|mappings]].
+Let $X = A \cup B$.
+Let $f$ and $g$ [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
+Then $f \cup g: X \to Y$ is a [[Definition:Mapping|mapping]].
+\end{theorem}
+
+\begin{proof}
+By definition, $f \cup g$ is a [[Definition:Relation|relation]] whose [[Definition:Domain of Relation|domain]] is $X = A \cup B$.
+Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$.
+At least one of the following must be true:
+:$(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$
+:$(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$
+:$(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$
+:$(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$
+Because $f$ and $g$ are [[Definition:Mapping|mappings]], $(1)$ and $(2)$ imply that $y_1 = y_2$.
+If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$.
+But then $x \in A \cup B$.
+So [[Definition:By Hypothesis|by hypothesis]]:
+:$y_1 = \map f x = \map g x = y_2$
+Similarly if $(4)$ holds.
+Thus in all cases:
+:$\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$
+and so by definition $f \cup g: X \to Y$ is a [[Definition:Mapping|mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Composite Mapping}
+Tags: Composite Mappings
+
+\begin{theorem}
+Let $S_1, S_2, S_3$ be [[Definition:Set|sets]].
+Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be [[Definition:Mapping|mappings]].
+Let $f_2 \circ f_1: S_1 \to S_3$ be the [[Definition:Composition of Mappings|composite mapping]] of $f_1$ and $f_2$.
+Then:
+:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
+where $\Dom {f_1}$ denotes the [[Definition:Domain of Mapping|domain]] of $f_1$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
+{{explain|Not quite what that definition says any more}}
+Let $x \in \Dom {f_2 \circ f_1}$.
+Then:
+:$\exists z \in S_3: \tuple {x, z} \in S_1 \times S_3$
+and:
+:$\exists y \in S_2: \tuple {x, y} \in f_1$
+That is:
+:$x \in \Dom {f_1}$
+Thus by definition of [[Definition:Subset|subset]]:
+:$\Dom {f_2 \circ f_1} \subseteq \Dom {f_1}$
+Now suppose $x \in \Dom {f_1}$.
+By the definition of [[Definition:Mapping|mapping]]:
+:$\exists y \in S_2: \tuple {x, y} \in f_1$
+As $f_2$ is likewise a [[Definition:Mapping|mapping]]:
+:$\exists z \in S_3: \tuple {y, z} \in f_2$
+and so by definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$\tuple {x, z} \in S_1 \times S_3$
+and so:
+:$x \in \Dom {f_2 \circ f_1}$
+Thus by definition of [[Definition:Subset|subset]]:
+:$\Dom {f_1} \subseteq \Dom {f_2 \circ f_1}$
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping is Idempotent}
+Tags: Identity Mappings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $I_S: S \to S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$.
+Then $I_S$ is [[Definition:Idempotent Mapping|idempotent]]:
+:$I_S \circ I_S = I_S$
+\end{theorem}
+
+\begin{proof}
+From [[Identity Mapping is Left Identity]]:
+:$I_S \circ f = f$
+for all [[Definition:Mapping|mappings]] $f: S \to S$.
+From [[Identity Mapping is Right Identity]]:
+:$f \circ I_S = f$
+for all [[Definition:Mapping|mappings]] $f: S \to S$.
+Substituting $I_S$ for $f$ in either one and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Functions Theorem}
+Tags: Mapping Theory, Named Theorems
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\set {X_i: i \in \N}$ be an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]] on $X$.
+For each $i \in \N$, let $g_i: X_i \to Y$ be a [[Definition:Mapping|mapping]] such that:
+:$g_{i + 1} \restriction X_i = g_i$
+where $g_{i + 1} \restriction X_i$ denotes the [[Definition:Restriction of Mapping|restriction]] of $g_{i + 1}$ to $g_i$.
+Then:
+:$\displaystyle \bigcup \set {g_i: i \in \N}$
+is a [[Definition:Mapping|mapping]] from $X$ to $Y$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\displaystyle g = \bigcup \set {g_i: i \in \N}$ is a [[Definition:Relation|relation]] whose [[Definition:Domain of Relation|domain]] is $X$.
+{{AimForCont}} $g$ is not a [[Definition:Mapping|mapping]].
+Then for some $x \in X$ and $i, h \in \N$:
+:$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$
+Let $k \in \N$ be the smallest such that:
+:$\map {g_i} x \ne g_{i + k}$
+where $x$ and $i$ are the same as in $(1)$.
+Then:
+:$\map {g_{i + k - 1} } x = g_i$
+But then:
+:$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$
+From this [[Proof by Contradiction|contradiction]] it follows that our supposition that $g$ is not a [[Definition:Mapping|mapping]] must be [[Definition:False|false]].
+That is: 
+:$\displaystyle \bigcup \set {g_i: i \in \N}$
+is a [[Definition:Mapping|mapping]] from $X$ to $Y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inductive Definition of Sequence}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $h \in \N$.
+Let $a_i \in X$ for all $i \in \set {1, 2, \ldots, h}$.
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Finite Sequence|finite sequences]] whose [[Definition:Codomain|codomains]] are in $X$.
+Let $G: S \to X$ be a [[Definition:Mapping|mapping]].
+Then there is a [[Definition:Unique|unique]] [[Definition:Sequence|sequence]] $f$ whose [[Definition:Codomain|codomain]] is in $X$ such that:
+:$f_i = \begin{cases}
+a_i & : i \in \set {1, 2, \ldots, h} \\
+\map G {f_1, f_2, \ldots, f_{i - 1} } & : i \ge h + 1
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+{{finish|tedious}}
+\end{proof}<|endoftext|>
+\section{Model of Root of Propositional Tableau is Model of Branch}
+Tags: Propositional Tableaus
+
+\begin{theorem}
+Let $\left({T, \mathbf H, \Phi}\right)$ be a [[Definition:Propositional Tableau|propositional tableau]].
+Let $v: \mathcal L_0 \to \left\{{T, F}\right\}$ be a [[Definition:Boolean Interpretation|boolean interpretation]] such that:
+:$v \models_{\mathrm{BI}} \mathbf H$
+that is, such that $v$ is a [[Definition:Model (Boolean Interpretations)|model]] for the [[Definition:Root of Propositional Tableau|root]] $\mathbf H$ of $T$.
+Then there exists a [[Definition:Branch (Graph Theory)|branch]] $\Gamma$ of $T$ such that:
+:$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$
+where $\Phi \left[{\Gamma}\right]$ denotes the [[Definition:Image of Subset under Mapping|image]] of $\Gamma$ under $\Phi$.
+\end{theorem}
+
+\begin{proof}
+We will find it convenient to reason with the [[Definition:Propositional Tableau/Construction|constructive definition of a propositional tableau]].
+First, let us reduce to the case of a finite [[Definition:Propositional Tableau|propositional tableau]].
+Suppose the result were to hold for these finite [[Definition:Propositional Tableau|tableaus]].
+Let $T$ be an infinite [[Definition:Propositional Tableau|propositional tableau]].
+Suppose that no [[Definition:Finite Branch|finite branch]] $\Gamma$ of $T$ satisfies $v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$.
+By definition, there exists an [[Definition:Exhausting Sequence of Sets|exhausting sequence]] $\left({T_n}\right)_{n \in \N}$ of finite tableaus for $T$.
+Define the [[Definition:Labeled Tree for Propositional Logic|labeled tree]] $T'$ as (the [[Definition:Induced Subgraph|subgraph induced]] by):
+:$t \in T'$ [[Definition:Iff|iff]] there exist an $n \in \N$, and a [[Definition:Branch (Graph Theory)|branch]] $\Gamma_n$ of $T_n$, such that $t \in \Gamma_n$ and $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_n}\right]$
+Suppose, aiming for a contradiction, that $T'$ is [[Definition:Finite Tree|finite]].
+By [[Branch of Finite Tree is Finite]], all [[Definition:Branch (Graph Theory)|branches]] of $T'$ are [[Definition:Finite Branch|finite]].
+By construction of $T'$, each of its [[Definition:Branch (Graph Theory)|branches]] $\Gamma$ satisfies:
+:$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$
+Because $T$ has no satisfactory [[Definition:Finite Branch|finite branch]], no such $\Gamma$ can be a [[Definition:Branch (Graph Theory)|branch]] of $T$.
+Hence for each [[Definition:Branch (Graph Theory)|branch]] $\Gamma$, there exists an $n \in \N$ such that $\Gamma \subseteq T_n$, but $\Gamma$ is no longer a [[Definition:Branch (Graph Theory)|branch]] of $T_n$.
+Since $T'$ is [[Definition:Finite Tree|finite]], it evidently has only finitely many [[Definition:Branch (Graph Theory)|branches]] $\Gamma_1, \ldots, \Gamma_m$.
+Now let $N \ge \max \left\{{n_i: 1 \le i \le m}\right\}$, with $n_i$ the least $n \in \N$ such that $\Gamma_i \subseteq T_n$ is not a [[Definition:Branch (Graph Theory)|branch]] of $T_n$.
+By definition of $T'$, this implies that there is no [[Definition:Branch (Graph Theory)|branch]] $\Gamma$ of $T_N$ such that:
+:$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$
+But this contradicts the assumption that the theorem holds for finite [[Definition:Propositional Tableau|propositional tableaus]].
+Hence it only remains to prove the theorem for finite [[Definition:Propositional Tableau|tableaus]].
+This we do by the [[Principle of Structural Induction]], as applicable to the definition of a [[Definition:Finite Propositional Tableau|finite propositional tableau]].
+In the remainder of this proof, "finite propositional tableau" is abbreviated to FPT for brevity.
+<section title="The case $\boxed{\mathrm{Root}}$" header="3" foldable="true">
+Let $T$ be a FPT identified by $\boxed{\mathrm{Root}}$.
+Then $T$ has the [[Definition:Branch (Graph Theory)|branch]] $\Gamma = \left\{{r_T}\right\}$ consisting only of the [[Definition:Root Node|root node]].
+Then $\Phi \left[{\Gamma}\right] = \mathbf H$, and by hypothesis on $v$:
+:$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\neg\neg}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\neg\neg}$ to another FPT $T_0$ at $t \in T_0$, using $\neg\neg\mathbf A$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$, which gets a [[Definition:Child WFF|child WFF]] $\mathbf A$.
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \mathbf A$.
+By [[Double Negation/Formulation 1/Proof 2|Double Negation]]:
+:$v \left({\mathbf A}\right) = v \left({\neg\neg\mathbf A}\right)$
+Since $\neg\neg\mathbf A$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\neg\neg\mathbf A \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that indeed $v \models_{\mathrm{BI}} \mathbf A$.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\land}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\land}$ to another FPT $T_0$ at $t \in T_0$, using $\mathbf A \land \mathbf B$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branch]] $\Gamma_r$ of the new [[Definition:Leaf Node|leaf node]], $r$, and it is seen that:
+:$\Phi \left[{\Gamma_r}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf A, \mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \left\{{\mathbf A, \mathbf B}\right\}$.
+Since $\mathbf A \land \mathbf B$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\mathbf A \land \mathbf B \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \mathbf A \land \mathbf B$.
+By the [[Rule of Simplification/Sequent Form/Formulation 1/Proof|Rule of Simplification]], we conclude:
+:$v \models_{\mathrm{BI}} \left\{{\mathbf A, \mathbf B}\right\}$
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\neg\land}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\land}$ to another FPT $T_0$ at $t \in T_0$, using $\neg \left({\mathbf A \land \mathbf B}\right)$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branches]] $\Gamma_s$ and $\Gamma_{s'}$ of the new [[Definition:Leaf Node|leaf nodes]], $s$ and $s'$, and it is seen that:
+:$\Phi \left[{\Gamma_s}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf A}\right\}$
+:$\Phi \left[{\Gamma_{s'}}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \neg\mathbf A$ or $v \models_{\mathrm{BI}} \neg\mathbf B$.
+Since $\neg \left({\mathbf A \land \mathbf B}\right)$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\neg \left({\mathbf A \land \mathbf B}\right) \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \neg \left({\mathbf A \land \mathbf B}\right)$.
+By the relevant [[De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Proof|De Morgan's Law]], it follows that:
+:$v \models_{\mathrm{BI}} \neg\mathbf A \lor \neg\mathbf B$
+which, by inspection of the [[Definition:Disjunction/Truth Table|truth table for $\lor$]], yields the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\lor}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\land}$ to another FPT $T_0$ at $t \in T_0$, using $\mathbf A \lor \mathbf B$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branches]] $\Gamma_s$ and $\Gamma_{s'}$ of the new [[Definition:Leaf Node|leaf nodes]], $s$ and $s'$, and it is seen that:
+:$\Phi \left[{\Gamma_s}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf A}\right\}$
+:$\Phi \left[{\Gamma_{s'}}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \mathbf A$ or $v \models_{\mathrm{BI}} \mathbf B$.
+Since $\mathbf A \lor \mathbf B$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\mathbf A \lor \mathbf B \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \mathbf A \lor \mathbf B$.
+By inspection of the [[Definition:Disjunction/Truth Table|truth table for $\lor$]], this establishes the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\neg\lor}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\neg\lor}$ to another FPT $T_0$ at $t \in T_0$, using $\neg \left({\mathbf A \lor \mathbf B}\right)$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branch]] $\Gamma_r$ of the new [[Definition:Leaf Node|leaf node]], $r$, and it is seen that:
+:$\Phi \left[{\Gamma_r}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf A, \neg\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \left\{{\neg\mathbf A, \neg\mathbf B}\right\}$.
+Since $\neg \left({\mathbf A \lor \mathbf B}\right)$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\neg \left({\mathbf A \lor \mathbf B}\right) \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \neg \left({\mathbf A \lor \mathbf B}\right)$.
+By the relevant [[De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Proof|De Morgan's Law]], it follows that:
+:$v \models_{\mathrm{BI}} \neg\mathbf A \land \neg\mathbf B$
+which, by inspection of the [[Definition:Conjunction/Truth Table|truth table for $\land$]], yields the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\implies}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\implies}$ to another FPT $T_0$ at $t \in T_0$, using $\mathbf A \implies \mathbf B$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branches]] $\Gamma_s$ and $\Gamma_{s'}$ of the new [[Definition:Leaf Node|leaf nodes]], $s$ and $s'$, and it is seen that:
+:$\Phi \left[{\Gamma_s}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf A}\right\}$
+:$\Phi \left[{\Gamma_{s'}}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \neg\mathbf A$ or $v \models_{\mathrm{BI}} \mathbf B$.
+Since $\mathbf A \implies \mathbf B$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\mathbf A \implies \mathbf B \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \mathbf A \implies \mathbf B$.
+By the [[Rule of Material Implication/Formulation 1/Proof|Rule of Material Implication]], it follows that:
+:$v \models_{\mathrm{BI}} \neg\mathbf A \lor \mathbf B$
+By inspection of the [[Definition:Disjunction/Truth Table|truth table for $\lor$]], this establishes the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\neg\implies}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\neg\implies}$ to another FPT $T_0$ at $t \in T_0$, using $\neg \left({\mathbf A \implies \mathbf B}\right)$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branch]] $\Gamma_r$ of the new [[Definition:Leaf Node|leaf node]], $r$, and it is seen that:
+:$\Phi \left[{\Gamma_r}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf A, \neg\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \left\{{\mathbf A, \neg\mathbf B}\right\}$.
+Since $\neg \left({\mathbf A \implies \mathbf B}\right)$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\neg \left({\mathbf A \implies \mathbf B}\right) \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \neg \left({\mathbf A \implies \mathbf B}\right)$.
+By [[Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Proof|Conjunction with Negative Equivalent to Negation of Implication]], we conclude:
+:$v \models_{\mathrm{BI}} \mathbf A \land \neg\mathbf B$
+which, by inspection of the [[Definition:Conjunction/Truth Table|truth table for $\land$]], yields the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\iff}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\iff}$ to another FPT $T_0$ at $t \in T_0$, using $\mathbf A \iff \mathbf B$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branches]] $\Gamma_s$ and $\Gamma_{s'}$ of the new [[Definition:Leaf Node|leaf nodes]], $s$ and $s'$, and it is seen that:
+:$\Phi \left[{\Gamma_s}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf A \land \mathbf B}\right\}$
+:$\Phi \left[{\Gamma_{s'}}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf A \land \neg\mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \mathbf A \land \mathbf B$ or $v \models_{\mathrm{BI}} \neg\mathbf A \land \neg\mathbf B$.
+Since $\mathbf A \iff \mathbf B$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\mathbf A \iff \mathbf B \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \mathbf A \iff \mathbf B$.
+By the [[Biconditional as Disjunction of Conjunctions/Formulation 1/Proof|Biconditional as Disjunction of Conjunctions]], it follows that:
+:$v \models_{\mathrm{BI}} \left({\mathbf A \land \mathbf B}\right) \lor \left({\neg\mathbf A \land \neg\mathbf B}\right)$
+By inspection of the [[Definition:Disjunction/Truth Table|truth table for $\lor$]], this establishes the desired result.
+{{qed|lemma}}</section>
+<section title="The case $\boxed{\neg\iff}$" header="3" foldable="true">
+Let $T$ be a FPT constructed by applying $\boxed{\neg\iff}$ to another FPT $T_0$ at $t \in T_0$, using $\neg \left({\mathbf A \iff \mathbf B}\right)$.
+By [[Leaf of Rooted Tree is on One Branch]], $t$ is on a unique [[Definition:Branch (Graph Theory)|branch]] $\Gamma_t$.
+The [[Definition:Branch (Graph Theory)|branches]] of $T$ are precisely those of $T_0$, except for $\Gamma_t$.
+$\Gamma_t$ is extended to the [[Definition:Branch (Graph Theory)|branches]] $\Gamma_s$ and $\Gamma_{s'}$ of the new [[Definition:Leaf Node|leaf nodes]], $s$ and $s'$, and it is seen that:
+:$\Phi \left[{\Gamma_s}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\mathbf A \land \neg \mathbf B}\right\}$
+:$\Phi \left[{\Gamma_{s'}}\right] = \Phi \left[{\Gamma_t}\right] \cup \left\{{\neg\mathbf A \land \mathbf B}\right\}$
+It will therefore suffice to show that:
+:If $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, then $v \models_{\mathrm{BI}} \mathbf A \land \neg\mathbf B$ or $v \models_{\mathrm{BI}} \neg\mathbf A \land \mathbf B$.
+Since $\neg\left({\mathbf A \iff \mathbf B}\right)$ is an [[Definition:Ancestor WFF|ancestor WFF]] of $t$, it follows that $\neg\left({\mathbf A \iff \mathbf B}\right) \in \Phi \left[{\Gamma_t}\right]$.
+Since $v \models_{\mathrm{BI}} \Phi \left[{\Gamma_t}\right]$, we conclude that $v \models_{\mathrm{BI}} \neg \left({\mathbf A \iff \mathbf B}\right)$.
+By the [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 2|Non-Equivalence as Disjunction of Conjunctions]], it follows that:
+:$v \models_{\mathrm{BI}} \left({\mathbf A \lor \neg \mathbf B}\right) \lor \left({\neg\mathbf A \land \mathbf B}\right)$
+By inspection of the [[Definition:Disjunction/Truth Table|truth table for $\lor$]], this establishes the desired result.
+{{qed|lemma}}</section>
+By the [[Principle of Structural Induction]], it follows that the result holds for every finite [[Definition:Propositional Tableau|propositional tableau]].
+As we had already reduced the theorem to the finite case, this completes the proof.
+{{qed}}
+\end{proof}
\ No newline at end of file