Prove $p \rightarrow \neg \neg p$ with Hilbert System

Asked Oct 16 '12 at 20:42

Active Oct 16 '12 at 20:42

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Possible Duplicate:
Prove that $\beta \rightarrow \neg \neg \beta$ is a theorem using standard axioms 1,2,3 and MP

I need to prove $p \rightarrow \neg \neg p$

My question is very similar to this, but with some differences. Axiom 3 that I'm using is different (though I've heard is equivalent) and the ONLY things I am allowed to use are the following three axioms, MP, and the Deduction Theorem:

A1: $A \rightarrow (B \rightarrow A)$

A2: $(A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))$

A3: $(\neg A \rightarrow \neg B) \rightarrow (B \rightarrow A)$

If another theorem/axiom is needed to solve this, it must be proven in-line of the main question. You cannot prove the extra theorem on the side and then use it in the proof.

edited Apr 13 '17 at 12:21

Community

asked Oct 16 '12 at 20:42

Kevin Cho

Also Proving (α → ¬¬α) in Hilbert System – MJD Oct 16 '12 at 21:32
I did reference the first link before, but it uses $\neg \neg p \rightarrow p$ as an already proven theorem. I would have to prove it within my proof so I wouldn't be able to use deduction theorem for it – Kevin Cho Oct 16 '12 at 23:17
The answer at http://math.stackexchange.com/a/21241/630 answers this too. As they point out, all that is needed are axiom schemes 1 and 2 and MP – Carl Mummert Oct 17 '12 at 02:49

Prove $p \rightarrow \neg \neg p$ with Hilbert System

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