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I have a question about intuitionistic logic regarding the relationship between the triple negation elimination rule, i.e. $\neg\neg\neg A\leftrightarrow \neg A$, and the double negation elimination. We know from Brouwer (1925) that $\neg\neg\neg A\leftrightarrow \neg A$ is true in intuitionistic logic. Let us define $B$ to be $\neg A$. Substitute $B$ and get $\neg\neg B\leftrightarrow B$. So does that mean the double negation elimination rule can still be applied in some cases in intuitionistic logic? Or am I missing something?

acevik
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  • It means you have double negation elimination for any $B$ that is already equivalent to $\neg A$ for some $A$. Of course not every $B$ necessarily has this. – halrankard Jul 19 '20 at 14:57

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does that mean the double negation elimination rule can still be applied in some cases in intuitionistic logic?

Yes, exactly right. Specifically, the double negation elimination rule $\lnot \lnot B \to B$ can be applied whenever $B$ is a negation, that is $B = \lnot A$ for some $A$, for the reasons you describe.

In general, many specific instances of double negation elimination are provable in intuitionistic logic. Another example is where $B$ is equivalent to $\bot$: then we can show that $\lnot \lnot B \to B$. But the statement $\lnot \lnot P \to P$ for a general predicate $P$ is not provable.

Caleb Stanford
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  • In fact, I've seen terminology (maybe "normal"?) for propositions satisfying $\lnot\lnot B \rightarrow B$. Some useful facts: both $\bot$ and $\top$ are normal; if $B$ is normal, then $A \rightarrow B$ is normal; as a corollary of the first two, $\lnot B$ is always normal; if $\phi$ (with possible free variable $x$) is normal, then $\forall x, \phi$ is normal. Though in the last case, you do need to be careful that in general, $\forall x, \lnot\lnot\phi$ is not necessarily equivalent to $\lnot\lnot\forall x, \phi$. – Daniel Schepler Aug 17 '20 at 21:36