TITLE: Use Fitch system to proof ((p ⇒ q) ⇒ p) ⇒ p without any premise. ONLY FOR FITCH SYSTEM.
QUESTION [6 upvotes]: I know here has few similar questions, but I cannot figure out with those answer. Since for Fitch system, I can only use And Intro, And Elim, Or Inro, Or Elim, Neg Intro, Neg Elim, Impl Intro, Impl Elim, Biconditional Intro, and Biconditional Elim.
I know I need to assume (p=>q)=>p then for next I need to prove p, at the end I can use Imlo Intro. But I don't know how to prove p...
I saw some answers like:
(p=>q)=>p  assume 1
p          assume 2
~p         assume 3
q          neg Elim with 2 or 3
I stop at this step..NOt sure why I can get q from 2 and 3? Since Neg Elim follows this form ( ~~p get p) and also Neg Intro follows (p=>q, p=>~q, get ~p). But from what I learn I can not get the q from just assume 2 and 3...
I tried in Fitch system, But it doesn't work with the answer from other questions. So that's right I post here again. Wish someone can help me.

REPLY [2 votes]: The issue with this problem seemed to be Stanford Fitch's approach to the negation rules, and their lack of a falsum symbol.

To prove the conditional $(p\to(p\to q))\to p$, assume $(p\to q)\to p$ aiming to derive $p$.  IE a conditional proof.
To derive $p$ from that assumption, assume $\lnot p$ aiming to derive a contradiction. Ie a proof by contradiction.
To derive a contradiction from $\lnot p$ and $(p\to q)\to p$, seek to derive $p\to q$.
To derive $p\to q$ from those assumptions, use a conditional proof. Ie: now assume $p$.
To derive $q$ from the assumptions of $p$, $\lnot p$, and $(p\to q)\to p$, well, we have assumed a contradiction so explode it.
Discharge the assumptions and we are done.
$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{}{\fitch{~~1.~(p\to q)\to p\hspace{5ex}\text{Assume}}{\fitch{~~2.~\lnot p\hspace{11ex}\text{Assume}}{\fitch {~3.~p\hspace{10ex}\text{Assume}}{~~4.~\bot\hspace{8ex}\lnot\,\mathsf E, 2,3\\~~5.~q\hspace{9ex}\text{EFQ}}\\~~6.~p\to q\hspace{7ex}\to\mathsf I,3{-}5\\~~7.~p\hspace{12ex}\to\mathsf E, 1,6\\~~8.~\bot\hspace{12ex}\lnot\,\mathsf E,2,7}\\~~9.~\lnot\lnot p\hspace{13ex}\lnot\,\mathsf I,2{-}8\\10.~p\hspace{16ex}\lnot\lnot\,\mathsf E, 9}\\11.~(p\to (q\to p))\to p\hspace{2ex}\to\mathsf I,1{-}10}$$
Now, the Stanford Fitch implimentation's of the negation introduction rule isn't too cumbersome (and what they call negation elimination is usually known as double negation elimination), but the lack of a falsum symbol means they do not have an explicit ex falso quodlibet rule (EFQ).  You have to use a proof by contradiction using their negation rules.   It bloats the proofs somewhat.
$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{}{\fitch{~~1.~(p\to q)\to p\hspace{5ex}\text{Assume}}{\fitch{~~2.~\lnot p\hspace{11ex}\text{Assume}}{\fitch {~3.~p\hspace{10ex}\text{Assume}}{\color{red}{\fitch{~~4.\lnot q\hspace{5ex}\text{Assume}}{~~5.~\lnot p\hspace{5ex}\mathsf R, 2}\\~~6.~\lnot q\to\lnot p\hspace{2ex}\to\mathsf I, 4{-}5\\\fitch{~~7.~\lnot q\hspace{5ex}\text{Assume}}{~~8.~p\hspace{7ex}\mathsf R, 3}\\~~9.~\lnot q\to p\hspace{3ex}\to\mathsf I, 7{-}8\\10.~~\lnot\lnot q\hspace{6ex}\lnot\,\mathsf I,6,9}\\11.~q\hspace{9ex}\lnot\lnot\,\mathsf E,10}\\12.~p\to q\hspace{7ex}\to\mathsf I,3{-}11\\13.~p\hspace{12ex}\to\mathsf E, 1,12}\\14.~\lnot p\to p\hspace{9ex}\to\mathsf E, 2{-}13\\\fitch{15.~\neg p\hspace{12ex}\text{Assume}}{}\\16.~\lnot p\to\lnot p\hspace{7.5ex}\to\mathsf I,15{-}15\\17.~\lnot\lnot p\hspace{13ex}\lnot\,\mathsf I,14,16\\18.~p\hspace{16ex}\lnot\lnot\,\mathsf E, 17}\\19.~(p\to (q\to p))\to p\hspace{2ex}\to\mathsf I,1{-}18}$$