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I have a feeling it's not, because ¬¬P → P is not provable. If it is, I'm not sure what kind of reductio I'd need to negate ¬(¬¬P → P). I believe a textbook somewhere said it was provable in intuitionistic logic, so am I missing something or is the textbook wrong?

debstack
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If is a propositional statement $\varphi$ is provable in classical logic, its double negation $\neg\neg\varphi$ is provable in intuitionistic logic. This fact is known as Glivenko's theorem (see e.g. here: https://en.wikipedia.org/wiki/Double-negation_translation).

The consequence is that your statement is provable in intuitionistic logic.

russoo
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    Double negation translation isn't just a matter of sticking $\lnot \lnot$ in front of the formula, though... – Zhen Lin Oct 23 '15 at 22:27
  • @ZhenLin: Ok, strictly speaking you are right. What I really had in mind was Glivenko's theorem. Fixed it! Thanks for your comment. – russoo Oct 23 '15 at 22:34
  • Thanks, this is just what I needed! Do you have a link somewhere where I can see the proof of Glivenko's theorem? I want to understand that first. – debstack Oct 23 '15 at 23:37
  • I am not able to find a clear exposition of the proof at the moment. But you can try to prove the theorem by yourself. If you are using Hilbert-style proof systems, you can proceed as follows: First, show that for each axiom $\varphi$ of the Hilbert system for classical logic, the formula $\neg\neg \varphi$ is provable in intuitionistic logic. Then, you should show that $\vdash_I \neg\neg \varphi$ and $\vdash_I \neg\neg(\varphi \to \psi)$ implies $\vdash_I \neg\neg\psi$ where $\vdash_I$ means "provable in intuitionistic logic". The result then follows by induction on the lengths of proofs. – russoo Oct 23 '15 at 23:50
  • A remark: For the above method of proof to work, it is sufficient if the classical logic is given by a Hilbert-style proof system. For the intuitionistic logic, you can use any kind of proof system you like (e.g. natural deduction for intuitionistic logic, tableaux systems for intuitionistic logic...). – russoo Oct 24 '15 at 00:04
  • Ah, I see. I have an idea of how the big picture of Glivenko's theorem would work. So I accept Glivenko's theorem and accept that ¬¬(¬¬P → P) is provable in intuitionistic logic. But I still don't see what the intuitionistic proof of ¬¬(¬¬P → P) would look like. Do you have an idea of what this would be? – debstack Oct 24 '15 at 00:20
  • The best thing is here that you simply try to prove your formula directly in the intuitionistic system you are working with. But an approximate idea of how the intuitionistic proof for your formula looks like can be inferred from the proof of Glivenko's theorem: Assume that we are using Hilbert systems for both, classical and intuitionistic logic. Construct a classical proof of $\neg\neg P \to P$. Now, double negate each formula of that proof. The resulting list is "almost" an intuitionistic proof for $\neg\neg(\neg\neg P \to P)$. More precisely, what the proof of Glivenko shows is ... – russoo Oct 24 '15 at 01:29
  • ... that the resulting list of double-negated formulas can always be completed in some uniform way into an intuitionistic proof of the formula in question. – russoo Oct 24 '15 at 01:32
  • Ah, I got it. Thanks! – debstack Oct 24 '15 at 16:37
  • @debstack Here is the proof you're looking for: http://math.stackexchange.com/questions/1584054/intuitionistic-proof-of-neg-neg-neg-neg-p-rightarrow-p/1584242 – Hans Brende Dec 21 '15 at 15:34
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The way to see this intuitively is: assume $\neg(\neg\neg P\implies P)$. Then assume $P$, by which $\neg\neg P \implies P$, a contradiction. Therefore $\neg P$. Then, if $\neg\neg P$, by ex falso $P$, so $\neg\neg P \implies P$, again a contradiction. Therefore $\neg\neg(\neg\neg P \implies P)$.

Tim Raczkowski
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Jan von Plato
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