prove this proposition is valid using the algebraic proposition

Question

Prove that tho proposition is valid using algebra

$$p \rightarrow q $$

$$q\rightarrow r$$

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$$\therefore \lnot p\lor r$$

what i've tried is

$\bigl((p\rightarrow q)\land(q\rightarrow r)\bigr)\rightarrow \lnot p\lor r$

$\lnot\bigl((\lnot p\lor q)\land(\lnot q\lor r)\bigr)\lor (\lnot p\lor r)$

and now im stuck

Answer 1

\begin{align*} (p \to q) \wedge (q \to r) \rightarrow (p \to r) &\Leftrightarrow (\lnot p \vee q) \wedge (\lnot q \vee r) \to (\lnot p \vee r) \\ &\Leftrightarrow \lnot((\lnot p \vee q) \wedge (\lnot q \vee r)) \vee (\lnot p \vee r) \\ &\Leftrightarrow (p \wedge \lnot q) \vee (q \wedge \lnot r) \vee \lnot p \vee r \\ &\Leftrightarrow (p \wedge \lnot q) \vee \lnot p \vee (q \wedge \lnot r) \vee r \\ &\Leftrightarrow ((p \vee \lnot p) \wedge (\lnot q \vee \lnot p)) \vee ((q \vee r) \wedge (\lnot r \vee r)) \\ &\Leftrightarrow \lnot q \vee \lnot p \vee q \vee r \\ &\Leftrightarrow \text{Tautology as $q \vee \lnot q$} \end{align*}