Prove $(\neg B \to \neg A) \to (A \to B)$ from axioms How can I prove that 
$$(\neg B \to \neg A) \to (A \to B),$$
if it is told that


*

*$A \to (B \to A),$ 

*$(A \to (B \to C)) \to ((A \to B) \to (A \to C)),$

*$(\neg B \to \neg A) \to ((\neg B \to \neg A) \to \neg B).$

 A: The second axiom needs some more parentheses. I suggest:
$$(A \implies (B \implies C)) 
\implies ((A \implies B) \implies (A \implies C))$$
More pressingly, your third axiom makes little sense. I am guessing you misstated it. I bet it should be:
$$(\neg B \implies A) \implies ((\neg B \implies \neg A) \implies B)$$
Finally, let's assume you are allowed the use of the Deduction Theorem.
Then we can do:


*

*$\neg B \implies \neg A$ Premise

*$A$ Premise

*$A \implies (\neg B \implies A)$ Axiom 1

*$\neg B \implies A$ MP 2,3

*$(\neg B \implies A) \implies ((\neg B \implies \neg A) \implies B)$ Axiom 3

*$(\neg B \implies \neg A) \implies B$ MP 4,5

*$B$ MP 1,6
Thus, we have shown $\neg B \implies \neg A, A \vdash B$
By the Deduction Theorem it thus follows that $\neg B \implies \neg A \vdash A \implies B$
And applying the Deduction Theorem on that, we get $\vdash (\neg B \implies \neg A) \implies (A \implies B)$
A: Alright, this problem as stated is impossible to solve.
Consider a model where $\lnot$ always returns a truth value of true, and ⟹ is the usual material conditional.  But, then all of the axioms hold true.  But, the conclusion can be false since if B is false and A is true, then ((¬B⟹¬A)⟹(A⟹B)) is false.
Maybe the third axiom got misstated?
