How to prove that every instance of the axiom $P\to(Q\to P)$ is a tautology?

Question

Bellow is what I tried. I think that's enough to prove that $P\to(Q\to P)$ is a tautology, but I'm not sure if that's enough to show that every instance of it is a tautology as well.

$P\to(Q\to P)$

$\equiv (\neg P \lor (Q\to P))$ (elimination of $\to$)

$\equiv (\neg P \lor (\neg Q \lor P))$ (elimination of $\to$)

$\equiv (\neg Q \lor (P \lor \neg P))$ (associativy)

$\equiv (\neg Q \lor T)$

$\equiv T$

Answer 1

Proceed by cases:

1. Assume $P$ is true.

Since $P$ is true, $Q → P$ is true by the truth table definition of "$\rightarrow$". Therefore, $P → (Q → P)$ is also true.

2. Assume $P$ is false.

Then $P → (Q → P)$ is true, again by the definition of "$\rightarrow$".

So $P → (Q → P)$ is true in either case. Therefore, it is a tautology.