Need to prove $(p \land q) \land (\lnot p \lor r) \rightarrow (q \lor r)$ is a tautology.

Question

I need to prove the following expression is a tautology using propositional logic laws.

My current working out is as follows [not sure if it is correct]:

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

Taking the LHS: $(p \land q) \land (\lnot p \lor r)$

$( (p \land q) \land \lnot p) \lor ( (p \land q) \land r)$ [Using distributive law]

$( (p \land \lnot p) \land q) \lor ( (p \land q) \land r)$ [Using associative law]

$( F \land q) \lor ( (p \land q) \land r)$ [Using complement law]

$F \lor ( (p \land q) \land r)$ [Using identity law]

This is where I get stuck. Is this the correct working so far? What other laws am I missing to prove the expression is a tautology?

Answer 1

$p \land q \land (\lnot p \lor r)$ implies $q$ $\tiny\text{… by simplification}$

$q$ implies $q \lor r$ $\tiny\text{… by addition}$

The rest is obvious.