Prove proposition using three axioms

Question

I need to prove that F → (∼G → ∼(F → G)) is universally valid, i.e. ⊨ F → (∼G → ∼(F → G)).

We are not allowed to use the Theorem of Completeness.

I tried going backwards and I got to F → ((F → G) → G), but I can't seem to get any closer to having only F.

If I try starting from F, none of the three axioms seem to help me get to F → ((F → G) → G).

The three axioms being :

1. F → (G → F)
2. (F → (G → H)) → ((F → G) → (F → H))
3. (∼F → ∼G) → (G → F)

Answer 1

In the comments you say that if you know how to get from $F$ to $(F\to G) \to G$, you'd be all set. OK:

\begin{array}{lll} 1. & F & Assumption\\ 2. & F \to ((F \to G) \to F) & Axiom \ 1\\ 3. & (F \to G) \to F & MP \ 1,2\\ 4. & (F \to G) \to (F \to G) & ID\\ 5. & ((F \to G) \to (F \to G)) \to (((F \to G) \to F) \to ((F \to G) \to G)) & Axiom \ 2\\ 6. & ((F \to G) \to F) \to ((F \to G) \to G) & MP \ 4,5\\ 7. & (F \to G) \to G & MP \ 3,6\\ \end{array}

Note that on line 4 I used ID (Identity). Most likely you have that one already, but if not, here is a proof:

\begin{array}{lll} 1. & P \to ((P \to P) \to P) & Axiom \ 1\\ 2. & (P \to ((P \to P) \to P)) \to ((P \to (P \to P)) \to (P \to P)) & Axiom \ 2\\ 3. & (P \to (P \to P)) \to (P \to P) & MP \ 1,2\\ 4. & P \to (P \to P) & Axiom \ 1\\ 5. & P \to P & MP \ 3,4\\ \end{array}