Demonstrate that (p → q) → ((q → r) → (p → r)) is a tautology.

Question

I'm struggling to demonstrate that

(p → q) → ((q → r) → (p → r))

is a tautology. I know that :

(p → q) → ((¬q ∨ r) → (¬p ∨ r))
        → (¬(¬q ∨ r) ∨ ¬(¬p ∨ r))
        → ((q ∧ ¬r) ∨ (p ∧ ¬r))     // De Morgan
        → ((q ∧ ¬r) ∨ (p ∧ ¬r))

I could also use distributivity to get to :

        → (¬r ∧ (p ∨ q))

Moving forward :

¬(p → q) ∨ (¬r ∧ (p ∨ q))
¬(¬p ∨ q) ∨ (¬r ∧ (p ∨ q))
(p ∧ ¬q) ∨ (¬r ∧ (p ∨ q))     // De Morgan

But I'm stuck there. Any help yould be appreciated !

Answer 1

Don't just apply Implication Equivalence to the last two implications, apply it to all four then apply DeMorgan's Laws and simplify.

$\begin{align} (p \to q) \to ((q \to r) \to (p \to r)) && \text{Given} \\ \neg (\neg p \lor q) \lor (\neg (\neg q \lor r) \lor (\neg p \lor r)) && \text{Implication Equivalence} \\ (p \land \neg q) \lor (q\land \neg r)\lor (\neg p \lor r)&& \text{DeMorgan's} \\ \neg p \lor (p \land \neg q) \lor r \lor (q\land \neg r)&& \text{Commuate and Associate} \\ \ddots & & \vdots \end{align}$