Logic and proofs: Prove that [tex](A \implies B) \implies \{ [(A \implies B)[/tex]...

[ksdhart2]

Hi all,

First off, I'm not quite sure which subforum to post this under, but I've chosen the "Advanced Math" because it doesn't seem to fit neatly in any of the other ones. I apologize in advance if I've messed up by doing so.

Anyway... I've recently started in a new math course involving logic and proofs, and I'm having some difficulty with a homework exercise. There was a set of three problems assigned, and I found two of them to be trivial, but the third is stumping me a bit. The problems are as follows:

2) Prove that \(\displaystyle (K \implies L) \implies [H \implies (K \implies L)]\) is a theorem

4) Prove that \(\displaystyle (A \implies B) \implies \{ [(A \implies B) \implies A] \implies A\}\) is a theorem

6) Prove that \(\displaystyle P \implies [(P \implies P) \implies P]\) is a theorem

For these exercises, we were given two axioms and we're not allowed to use anything else unless we first prove it ourselves. These two axioms are:

Axiom 1, Law of Affirmation of the Antecedent: \(\displaystyle \vdash P \implies (Q \implies P)\)

Axiom 2, Law of Transitivity of the Implication: \(\displaystyle \vdash [P \implies (Q \implies R)] \implies [(P \implies Q) \implies (P \implies R)]\)

I've successfully completed problems #2 and #6. For #2, I substituted \(\displaystyle (K \implies L)\) for \(\displaystyle P\) and \(\displaystyle H\) for \(\displaystyle Q\) in Axiom 1, directly yielding the desired outcome in one step. Similarly, for #6, I substituted \(\displaystyle (P \implies P)\) for \(\displaystyle Q\) in Axiom 1, again directly yielding the desired outcome in one step.

However, problem #4 is giving me troubles. I believe I know a good strategy. The book gives this theorem:

\(\displaystyle [H \implies (K \implies L)] \implies [K \implies (H \implies L)]\)

I think I can prove that theorem and then use it to jump from:

\(\displaystyle \vdash [(A \implies B) \implies A] \implies [ (A \implies B) \implies A]\) to
\(\displaystyle \vdash (A \implies B) \implies \{ [(A \implies B) \implies A] \implies A\}\)

but I'm uncertain as to how I might arrive at that first part in order to do so, starting from just the two axioms. Any help on this problem would be greatly appreciated.