Derive formal or informal proofs using Modus Ponus and Deduction Theorem - Logic

Question

How do I show the following by producing formal or informal proofs?

1. ⊢ (¬A → ¬B) → (B → A )

2. ⊢ ¬B → (B → A)

I can use the Modus ponus (MP) rule, and deduction theorem (DT). And I have these 3 axioms:

1. α → (β → α) --- (A1)

2. (α → (β → γ)) → ((α → β) → (α → γ)) --- (A2)

3. (¬β → ¬α) → ((¬β → α) → β) --- (A3)

Thank you!

Answer 1

Hint:

1) With DT we firstly conclude that it is enough to prove $\left\{ \neg A\to\neg B\right\} \vdash B\to A$ and based on that secondly that it is enough to prove $\left\{ \neg A\to\neg B,B\right\} \vdash A$.

2) With DT we firstly conclude that it is enough to prove $\left\{ \neg B\right\} \vdash B\to A$ and based on that secondly that it is enough to prove $\left\{ \neg B,B\right\} \vdash A$.

Now give that a try and tell us where you get stuck in the sequel.

Answer 2

For the first one:

$1\ \neg A \to \neg B \ Premise$

$2 \ B \ Premise$

$3 \ B \to (\neg A \to B) \ Axiom \ 1$

$4 \ \neg A \to B \ MP \ 2,3$

$5 \ (\neg A \to \neg B) \to ((\neg A \to B) \to A \ Axiom \ 3$

$6 \ (\neg A \to B) \to A \ MP \ 1,5$

$7 \ A \ MP \ 4,6$

This shows that $\{ \neg A \to \neg B, B \vdash A\}$

So, with two applications of DT, you're there.