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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 $?

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  • $\begingroup$ The distributive property might be helpful: $(a\vee(b\wedge c))\equiv((a\vee b)\wedge(a\vee c))$ $\endgroup$
    – 79037662
    Sep 26, 2019 at 13:46

2 Answers 2

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

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

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

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

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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 $p\vee q \implies p\vee q\implies (p\vee q)\land(q\land r)$

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