TITLE: Does this variant of Peirce's law hold in intuitionistic logic? $\vdash ((A \rightarrow B) \rightarrow A) \implies \vdash A$
QUESTION [5 upvotes]: In intuitionistic logic, Peirce's Law, $((A \rightarrow B) \rightarrow A) \rightarrow A$, does not hold.
However, I am wondering about a slight modification to this same law:
$$\vdash ((A \rightarrow B) \rightarrow A) \text{ implies } \vdash A$$
i.e., if $(A \rightarrow B) \rightarrow A$ is provable, then $A$ is also provable, for any values of $A$ and $B$.
My question is: does this meta-version of Peirce's Law hold in intuitionistic logic? I'm asking this question after trying unsuccessfully to find any counter-examples, but also not knowing how to prove the validity of such a statement.
I'm looking for either (1) a proof that it holds, or (2) a counter-example (showing, for some $A$ and $B$, an intuitionistic proof of $(A \rightarrow B) \rightarrow A$, where $A$ is not provable in intuitionistic logic).

REPLY [2 votes]: I've realized that a counter-example exists, so my modification to Peirce's law does not hold in intuitionistic logic.
We must find some $A$ and $B$ such that $A$ is not provable in intuitionistic logic, while $(A \to B) \to A$ is provable in intuitionistic logic.
Let $A = \neg \neg P\to P$, and let $B = \neg P$
$A$ is clearly not provable in intuitionistic logic, since it is equivalent to the double negation elimination rule, which does not hold intuitionistically.
It remains to prove: $((\neg\neg P \to P)\to\neg P) \to (\neg\neg P \to P)$

Suppose $(\neg\neg P \to P)\to\neg P$  
Since it's tautologically true that $P\to (\neg\neg P \to P)$, we can deduce that $P\to\neg P$. Therefore, $\neg P$.
Now suppose $\neg\neg P$. But this contradicts $\neg P$. Therefore, by the principle of explosion, we can say: $\neg\neg P \to P$. 
Therefore, $((\neg\neg P \to P)\to\neg P) \to (\neg\neg P \to P)$, Q.E.D.

Since this proof requires the principle of explosion to work, I believe my modified version of Peirce's law would still hold in minimal logic.
Edit: It doesn't hold in minimal logic either. Proof:
Let $A = ((P\to Q)\to P)\to P$, let $B = P\to Q$.
$A$, being equivalent to Peirce's Law, is clearly not provable in minimal logic.
It remains to prove: $((((P\to Q)\to P)\to P)\to (P\to Q))\to (((P\to Q)\to P)\to P)$

Suppose $(((P\to Q)\to P)\to P)\to (P\to Q)$.
Since it's tautologically true that $P\to (((P\to Q)\to P)\to P)$, we can deduce from (1) that $P\to(P\to Q)$, and hence that $P\to Q$. 
Suppose, in addition, that $(P\to Q)\to P$. Then we can deduce $P$, by modus ponens from (2). Therefore, $((P\to Q)\to P)\to P$.
Therefore, $((((P\to Q)\to P)\to P)\to (P\to Q))\to (((P\to Q)\to P)\to P)$, Q.E.D.