$[(p \to q) \land (q \to r)] \to (p \to r)$ is a tautology

Question

I am trying to prove, using logical equivalences, that $$[(p \to q) \land (q \to r)] \to (p \to r)$$ is a tautology. My attempt:

$$\sim[(p\to q) \land (p\to r) ] \lor(\sim p \lor r) \\ \equiv [\sim (p\to q) \lor \sim (q\to r) ]\lor (\sim p \lor r) \\ \equiv [(p\land \sim q)\lor (q \land \sim r ) ]\lor (\sim p \lor r).$$

Answer 1

Let $p$, $q$, and $r$ be given propositions. Assume that $p$ implies $q$ and $q$ implies $r$ are true statements. If $p$ is false, then the statement $p$ implies $r$ is true. Suppose from now on that $p$ is true. By assumption, $q$ is true and then $r$ is true. We have proved that $p$ implies $r$ is true. Now can you translate this into symbols?

Answer 2

Expanding your final line mechanically by distribution laws,

$$\begin{align*} & [(p\land \neg q) \vee (q\wedge \neg r )] \vee \neg p \vee r\\ &= [(p\vee q)\wedge (p\vee\neg r) \wedge (\neg q \vee q) \wedge (\neg q \vee \neg r)] \vee \neg p \vee r\\ &= (p\vee q \vee \neg p \vee r)\wedge (p\vee\neg r \vee \neg p \vee r) \wedge (\neg q \vee q \vee \neg p \vee r) \wedge (\neg q \vee \neg r \vee \neg p \vee r)\\ &= (p\vee \neg p \vee \cdots)\wedge (p\vee \neg p \vee \cdots) \wedge (\neg q \vee q \vee \cdots) \wedge (\neg r \vee r\vee \cdots)\\ &= T \wedge T \wedge T \wedge T\\ &= T \end{align*}$$