I'm trying to show that $\vdash \neg\neg A \to A$. I can't seem to figure out the deduction. Mendelson proves this in his book, but I'm trying to use a different set of axioms. These are
$A\to (B\to A)$
$(\neg B\to \neg A)\to (A\to B)$
$(A\to (B\to C))\to ((A\to B)\to (A\to C))$
Using these axioms and modus ponens I've done a few problems in Mendelson's book, so I already have deductions for e.g.
$\neg A \to (A\to B)$
Can anyone find a short proof for this? Even using the deduction theorem, I can't seem to be able to prove this.