TITLE: If $\Gamma$ is consistent and $\Gamma\not\vdash\phi$, then $\Gamma\cup\{\neg\phi\}$ is also consistent. Why?
QUESTION [6 upvotes]: I read this answer, but was a bit confused by one of the deductions. It is stated that for some first-order theory $\Gamma$ and statement $\phi$:

If $\Gamma$ is consistent and $\Gamma\not\vdash\phi$, then $\Gamma\cup\{\neg\phi\}$ is also consistent.

This looks natural at first, and the author stated that this is by definition. But for me it is far from obvious that $\neg\phi$ cannot be used to deduce $\phi$ when combined with the other axioms in $\Gamma$. Can someone explain? Am I missing some definition or an obvious point?

REPLY [2 votes]: Note that the above argument use nonconstructive law of reductio ad absurdum:
Let $X\cup\{\neg\varphi\}$ be an inconsistent set of formulas. Then $X\cup\{\neg\varphi\}\vdash\varphi $, so Deduction Theorem gives: $X\vdash\neg\varphi\rightarrow\varphi $. But reductio ad absurdum  $(\neg\varphi\rightarrow\varphi)\rightarrow \varphi $ is clasically valid, so by Modus Ponens, $X\vdash\varphi $. $\blacksquare$
This theorem is not true in Intuitionistic Logic. Indeed, $\not\vdash_{INT}p\vee\neg p$, but $\{\neg(p\vee\neg p)\}$ is inconsistent:
suppose that there is a finite Kripke model $K$, and a world $x\in K$, such that $x\Vdash \neg(p\vee\neg p)$. Then taking a maximal element $y$ over $x$, we have $y\not\Vdash p\vee\neg p$, which is mpossible. $\blacksquare$ 
In Intuitionistic Logic the weaker form holds: If  $X\cup\{\varphi\}$ is inconsistent, then $X\vdash_{INT}\neg\varphi $.