Super Simple question on Logic and Modus Ponens I am totally mixed up with these:
using ONLY this three axioms and Modus Ponens:$$1. \ F \implies (G\implies F) \\ 2. \ (F \implies (G\implies H))\implies ((F \implies G)\implies (F \implies H)) \\
3. \ (\neg G \implies \neg F)\implies ((\neg G\implies F)\implies G)$$
I need to prove:
$$4. \ (\neg G\implies \neg F) \implies (F \implies G)\\
5. (F \implies G)\implies ((G \implies H)\implies(F\implies H))\\
6.(F\implies (G \implies H))\implies(G\implies (F\implies H)$$
I see it intuitively, but I have to use ONLY the axioms.
I appreciate any help.
 A: In oder to prove 5 and 6, we need some preliminary results.

T1 : $P \rightarrow P$
1) $P \rightarrow ((Q \rightarrow P) \rightarrow P)$ --- Ax.1
2) $P \rightarrow (Q \rightarrow P)$ --- Ax.1
3) $(1) \rightarrow ((2) \rightarrow (P \rightarrow P))$ --- Ax.2

4) $P \rightarrow P$ --- from 3), 1) and 2) by Modus Ponens twice.


T2 : $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$
1) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2
2) $(1) \rightarrow ((Q \rightarrow R) \rightarrow (1))$ --- Ax.1
3) $(Q \rightarrow R) \rightarrow (1)$ --- from 1) and 2) by Modus Ponens
4) $(Q \rightarrow R) \rightarrow (P \rightarrow (Q \rightarrow R))$ --- Ax.1
5) $(3) \rightarrow ((4) \rightarrow ((Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))))$ --- Ax.2

6) $(Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- from 5), 3) and 4) by MP twice.


T3 : $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$
1) $(P \rightarrow Q) \rightarrow (P \rightarrow Q)$ --- T1
2) $(1) \rightarrow (((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q))$ --- Ax.2
3) $((P \rightarrow Q) \rightarrow P) \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 1) and 2) by MP
4)  $P \rightarrow ((P \rightarrow Q) \rightarrow P)$ --- Ax.1
5) $(3) \rightarrow ((4) \rightarrow (P \rightarrow ((P \rightarrow Q) \rightarrow Q)))$   --- T2

6) $P \rightarrow ((P \rightarrow Q) \rightarrow Q)$ --- from 5), 3) and 4) by MP twice.


T4 : $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$
1) $((P \rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow ((Q \rightarrow (P \rightarrow Q)) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- T2
2) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ --- Ax.2
3) $Q \rightarrow (P \rightarrow Q)$ --- Ax.1
4) $(1) \rightarrow ((2) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))))$ --- T2
5) $(P \rightarrow (Q \rightarrow R)) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R)))$ --- from 4), 1) and 2) by MP twice
6) $(3) \rightarrow ((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- T3
7) $((3) \rightarrow (Q \rightarrow (P \rightarrow R))) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 6) and 3) by MP
8) $(7) \rightarrow ((5) \rightarrow ((P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))))$ --- T2

9) $(P \rightarrow (Q \rightarrow R)) \rightarrow (Q \rightarrow (P \rightarrow R))$ --- from 8), 7) and 5) by MP twice.

This is 6 above.

T5 : $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$
1) $((Q \rightarrow R) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))) \rightarrow ((P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R)))$ --- T4

2) $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$ --- from 1) and T2 by MP.

This is 5 above.
A: To prove 4...
Step 1.) Suppose $\neg G\Longrightarrow\neg F$.  (We wish to finally conclude $F\Longrightarrow G$.)
Step 2.) Then $((\neg G\Longrightarrow F)\Longrightarrow G$ (by $(3)$).
Step 3.) Suppose now $F$.
Step 4.) Then $\neg G\Longrightarrow F$ (by Step $3$ and $(1)$).
Step 5.) Then $G$ (by Step (4), Step $(2)$, and Modes Ponens).
Step 6.) Then $F\Longrightarrow G$ (by Step $3$ and Step $5$).
Conclude finally that $(\neg G\Longrightarrow\neg F)\Longrightarrow(F\Longrightarrow G)$ (by Step $1$ and Step $6$).
