I've used the rules of Chapter 2 of van Dalen's Logic and Structure, which allows only $\{ \wedge I, \wedge E, \to I, \to E, \bot E, RAA.\}$ $$ \newcommand{\bfrac}[2]{\displaystyle\genfrac{}{}{0pt}{}{#1}{#2}} \cfrac{ \cfrac{ \cfrac{{\bfrac{}{[( P\to Q)\to P]^3}} \quad\quad\quad \cfrac{[Q]^1}{P\to Q}}{P} \quad\quad\quad \bfrac{}{ {[\neg P]^2} }} {{\cfrac{\cfrac{\bot}{ \neg Q}1 \quad\quad\quad\quad\quad \bfrac{}{[ Q] ^1}} {\cfrac{\bot}{P}}}}} {(P\to Q)\to P)\to P}3 $$
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Please review your use of MathJax – Sep 03 '21 at 06:31
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Are you sue this is how it is supposed to be written? It's a tad bit hard to read. – Sep 03 '21 at 06:34
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3No; the "lower" assumption of $Q$ is not correctly discharged. – Mauro ALLEGRANZA Sep 03 '21 at 06:39
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1I've just edited. mathjax doesn't support proof tree, so i can't choose another way to proof except the method to use cfrac. – blahblah Sep 03 '21 at 06:40
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1See this post: you need Double Negation. – Mauro ALLEGRANZA Sep 03 '21 at 06:40
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Thank you, Mauro ALLEGRANZA. – blahblah Sep 03 '21 at 06:40
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Yes, now you derivation is perfect. Just add 2 next to the RAA inference rule deriving $P$ and discharging $\lnot P$, so that it's explicit where the assumption $\lnot P$ has been discharged. – Taroccoesbrocco Sep 03 '21 at 08:12
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I think the following is the most natural way for me to derive Peirce's law in classic logic per Dalen's natural deduction rules and style. Another style is Fitch which I found in an old post here deriving the same Peirce's law in Fitch style, and the steps are essentially the same as mine below.
Please note that in your derivation you need to assume $Q$ (your assumption #1) which is not needed at all in the whole proof. Actually, if you transform your own proof to Fitch, then you'll see you cannot arrive at your first $\bot$ above at all within the same subproof. IMHO, Dalen's style is error-prone and not suitable for new students just beginning their logic study. You seem to be able to do the job by discharging all assumptions, but you also intuitively feel you're not confident at all.
$$ \cfrac{\lower{2ex}{[\lnot P]^2} \quad \cfrac { \lower{2ex}{[(P \to Q) \to P]^1} \quad \cfrac { \cfrac { \cfrac {[\neg P]^2 \quad [P]^3} {\bot} \to E } {Q} \bot } {(P \to Q)} \to I_3 } { P } \to E } { \cfrac { \cfrac { \bot } {P} \quad RAA_2 } {((P \to Q) \to P) \to P} \to I_1 } \to E $$
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