# How to prove this propositional tautology using only axioms from Mendelson's “Introduction to Mathematical Logic”

The result I wish to prove is (A -> (B -> C)) -> (B -> (A -> C))

Firstly does this have a name? I've been calling it "Swapping Hypothesis".

Secondly, I'm trying to find a proof of this only using the axioms from the book "Introduction to Mathematical Logic" by Mendelson. He gives three propositional axioms, here are 2 of those axioms

• A1 (A -> (B -> A)
• A2 ((A -> (B -> C)) -> ((A -> B) -> (A -> C))

The rule of proof is Given A and A -> B then B

The third axiom I do not believe is relevant.

I've already proved a couple of results, which may be of use:

• (Theorem) (B -> C) -> ((A -> B) -> (A -> C))
• (Rule) Given (A -> B) and (B -> C) then (A -> C) (Hypothetical Syllogism)

If you can do it without Deduction Theorem that would be great, but it's OK with Deduction also (as I think I can translate a proof with Deduction into a proof without fairly easily).

https://math.stackexchange.com/a/1071904/123948

• The 'law of commutation' is one of the names. – Doug Spoonwood May 21 at 13:03
• Have you proved the Deduction Theorem? Then it is very straightforward. – Henning Makholm May 21 at 14:32
• If the third axiom is $(\neg A\to\neg B)\to(B\to A)$ then it does actually have new consequences that can be written using only $\to$ -- such as Pierce's law $((P\to Q)\to P)\to P$. But what you're aiming for here doesn't need it. – Henning Makholm May 21 at 14:35
• Thanks @HenningMakholm I'll update my question. – samthebest May 21 at 14:41

The main heuristic is: Each time you need to prove something of the form $$\cdots\to\cdots$$, apply the Deduction Theorem. Once that won't get you any further, you'll have assumed $$A\to(B\to C)$$ and $$B$$ and $$A$$, and you need to prove $$C$$. But that is then just two applications of modus ponens away.