Proof logical equivalence without truth-table

Question

Prove without using table truth the following equivalence:

$(p \, \vee q) \,\vee (q \, \land r) \Leftrightarrow p \, \vee q $

Some light on how to "eliminate" this $ r $?

Answer 1

$(p \vee q) \vee (q \wedge r)$

$\Leftrightarrow p \vee (q \vee (q \wedge r))$ by the associative law

$\Leftrightarrow p \vee q$ by the absorption law

Hence, $(p \vee q) \vee (q \wedge r) \Leftrightarrow p \vee q$

Answer 2

If $p\vee q$, obviously, $p\vee q$. Otherwise, if $q\land r\implies q\implies p\vee q$. So, the $\implies$ is ok.

On the other hand, $p\implies p\vee q$ and $q\implies p\vee q$, so $p\vee q \implies p\vee q\implies (p\vee q)\land(q\land r)$