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We can conclude $\neg\neg B$ from $A \rightarrow B$ and $\neg A \rightarrow B$ in intuitionistic propositional logic.

More precisely, in second-order propositional logic, $\neg\neg B$ is in fact equivalent to $\exists A. (A \vee \neg A) \rightarrow B$.

Question: How does one prove this?

Motivation: This question comes up surprisingly frequently online, most recently here. Common contexts involve proofs of Littlewood's theorem that invoke excluded middle on the Riemann hypothesis. I am self-answering it to have an answer I can reference easily when it comes up again.

Z. A. K.
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We can prove the non-second-order version as follows:

Given the assumption $A \rightarrow B$, we can obtain $\neg B \rightarrow \neg A$ by the intuitionistically valid direction of taking contrapositives (see here).

If we also have $\neg A \rightarrow B$, transitivity of implication gets us to $\neg B \rightarrow B$, which is equivalent to $\neg\neg B$.

Alternative proof of this first claim: given $A\rightarrow B$ and $\neg A \rightarrow B$, we have $(A \vee \neg A) \rightarrow B$, and therefore $\neg\neg(A \vee \neg A) \rightarrow \neg\neg B$ using the fact that double negation distributes over implication (two uses of contraposition). But $\neg\neg(A \vee \neg A)$ is provable by Glivenko's theorem, so $\neg\neg B$ follows.

For the second-order version, we can argue as follows:

If there's some proposition $A$ such that $(A \vee \neg A) \rightarrow B$, then the previous argument lets us conclude $\neg\neg B$ again. This takes care of one direction.

For the other direction, assume $\neg\neg B$ holds. We need to prove that there's some $A$ such that assuming $A \vee \neg A$ lets us conclude $B$.

But setting $A:=B$ suffices. If we assume $B \vee \neg B$, then we can argue by cases. In the first case, $B$ holds so we're done. In the second case, $\neg B$ holds, which contradicts $\neg\neg B$: but everything follows from a contradiction, including $B$. In either case, $B$ follows, so we have managed to prove $(B\vee \neg B)\rightarrow B$, and hence $\exists A. (A \vee \neg A)\rightarrow B$, from the assumption $\neg\neg B$.

Z. A. K.
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    Yet another minor variant of the non-second-order version: from the hypotheses, you can conclude $\lnot B \rightarrow \lnot A$ and $\lnot B \rightarrow \lnot\lnot A$. Therefore, from $\lnot B$ you can conclude both $\lnot A$ and $\lnot \lnot A$ which together give a contradiction, showing $\lnot \lnot B$. – Daniel Schepler Nov 06 '24 at 17:47
  • It has been a few months, so I'll accept this now take it off the list of people who filter for "no accepted answers". As customary with questions asked only to be self-answered, as soon as somebody else posts a better or substantially different answer, I'll accept that instead. – Z. A. K. May 01 '25 at 01:22