Fitch proof of Peirce's Law

Asked Oct 05 '18 at 00:04

Active Oct 05 '18 at 00:13

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This was a question on my exam today that I couldn't figure out.

Show $((P \supset Q) \supset P) \supset P$ with no premises using Fitch.

The first attempt is I assumed $(P \supset Q) \supset P$. Then I assume $\neg P$ and attempted to derive a contradiction.

However, I couldn't get anywhere productive. Any hints or direction?

edited Oct 05 '18 at 00:13

asked Oct 05 '18 at 00:04

1

Hint: under the assumption $\lnot P$, it follows that $P \supset Q$ for any $Q$. – Daniel Schepler Oct 05 '18 at 00:07
@DanielSchepler Could you lead me to that? I see that using Modus Tollens we can get $\neg(P \supset Q)$. But I don't think that's exactly what you were talking about. – Good Morning Captain Oct 05 '18 at 00:11
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Well, if you have assumptions $\lnot P, P$ then you can conclude $\bot$ and therefore by ex falso quodlibet (explosion principle) $Q$ follows. So, then by ${\supset} I$ with the assumption $\lnot P$ you conclude $P \supset Q$. – Daniel Schepler Oct 05 '18 at 00:13
@DanielSchepler Oh oh oh. Okay. I appreciate it. – Good Morning Captain Oct 05 '18 at 00:18

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