How do i prove that $(p \vee q) \wedge (\neg p \vee r) \rightarrow (q \vee r)$ is a tautology without using the truth table?

Question

I am looking to for a way to prove that the statement $(p \vee q) \wedge (\neg p \vee r) \rightarrow (q \vee r)$ is a tautology. I am unable to use the truth table or the rules of inference for this. I can only solve this by using Laws and Theorems such as De Morgans Law and Distributive Law. Any help would be greatly appreciated. Thank you

Answer 1

Assume $\lnot(q \vee r)$ and reach a contradiction. You may have to make a second level assumption once of $p$ and once of $\lnot p$ and reach a contradiction from each one.

Answer 2

\begin{align} (p \lor q) \land (\neg p \lor r) \to (q \lor r)&\equiv\neg\{[(p \lor q) \wedge (\neg p \lor r)] \land \neg(q \lor r)\}\\&\equiv\neg[(p\lor q)\land(\neg p\lor r)\land(\neg q\land\neg r)]\\&\equiv\neg[(p\lor q)\land\neg p\land\neg q\land\neg r]\\&\equiv\neg[(p\lor q)\land\neg(p\lor q)\land \neg r)]\\&\equiv\neg0\equiv1 \end{align}