I'm having trouble understanding how to go about coming up with these proofs. I understand that we have a list of axioms and a reference rule (Modus Ponens) to make use of and we have to use a combination of them to get down to what we are trying to prove.

$\displaystyle \vdash ((\neg p \rightarrow) \rightarrow p)$

I tried this:

$\displaystyle (\neg p \rightarrow p) \rightarrow ((\neg p \rightarrow \neg p) \rightarrow p) (by Ax3)$

$\displaystyle (\neg p \rightarrow p) \rightarrow (\neg p \rightarrow p) (by MP, 1)$

$\displaystyle (\neg p \rightarrow p) \rightarrow p (by MP, 2, 1)$

But this doesn't seem right to me. It seems way too short. So what I'm confused about is how to actually go about these proofs. I'm still confused by what is meant by Modus Ponens.

That was if $\displaystyle A \rightarrow B$ and A then B. Does this just mean if I have $\displaystyle A \rightarrow B$ that I can replace that with just B?