Can Modus Ponens and Substitution Solve This Complex Logic Problem?

[JustNick]

Using only modus ponens or substitution. Prove:
q -> r -> [ [ p -> q ] -> [ p -> r ] ]

using the three axioms:
1) p -> [ q -> p ]
2) s -> [ p-> q ] -> [ [s -> p] -> [ s -> q ] ]
3) p -> f -> f -> p

where the symbol f is "false."

I am having the hardest time trying to solve this proof, any point in the right direction is appreciated.

 

[fresh_42]

If p then it follows from 2). (s≡p ; p≡q ; q≡r) But isn't 3) just p? I guess my understanding of 3) is f.
May I conclude f → (p → f) by 1) and then f → p by 3)? If so I'm done by 2) and the substitutions mentioned.