Prove $(\neg B \to \neg A) \to (A \to B)$ from axioms

Question

How can I prove that $$(\neg B \to \neg A) \to (A \to B),$$

if it is told that

1. $A \to (B \to A),$
2. $(A \to (B \to C)) \to ((A \to B) \to (A \to C)),$
3. $(\neg B \to \neg A) \to ((\neg B \to \neg A) \to \neg B).$

Answer 1

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:

1. $\neg B \implies \neg A$ Premise

2. $A$ Premise

3. $A \implies (\neg B \implies A)$ Axiom 1

4. $\neg B \implies A$ MP 2,3

5. $(\neg B \implies A) \implies ((\neg B \implies \neg A) \implies B)$ Axiom 3

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

7. $B$ 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 $\vdash (\neg B \implies \neg A) \implies (A \implies B)$

Answer 2

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?