Prove that ¬P → ( P → ( P → Q)) is a tautology without using truth tables

Question

I can't find a proper formula for this considering it's almost exclusively made up of implications. Can somebody help me?

EDIT: Sorry, I'm new to this site and still learning to use it. I've tried writing (P → Q) as (¬P ∨ Q) and then applying the distibutive laws but I feel like I've reached a dead end.

Answer 1

P -> q is the same as no(p) OR q If you replace, in your expression :

P -> (P -> Q) is the same as no(P) OR (no(P) OR Q)

no(P) -> P (P -> (P -> Q)) is the same as no(no(p)) OR (no(P) OR (no(P) OR Q)) which is the same as p OR no(P) OR no(P) OR Q which is always true ( because p or no(p) is always true)

Answer 2

Simply:

!P -> (P -> (P -> Q))  apply implication
P v (!P v (P -> Q)))   P v !P is T
T v (...)              T v anything is T
T