Using truth tables, prove the following logical equivalence :
$p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$

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Solution

$p$ $q$ $r$ $q \land r$ $p \lor q$ $p \lor r$ $p \lor (q \land r)$ $(p \lor q) \land (p \lor r)$ Truth
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$ True
$T$ $T$ $F$ $F$ $T$ $T$ $T$ $T$ True
$T$ $F$ $T$ $F$ $T$ $T$ $T$ $T$ True
$T$ $F$ $F$ $F$ $T$ $T$ $T$ $T$ True
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$ True
$F$ $T$ $F$ $F$ $T$ $F$ $F$ $F$ True
$F$ $F$ $T$ $F$ $F$ $T$ $F$ $F$ True
$F$ $F$ $F$ $F$ $F$ $F$ $F$ $F$ True

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Similar questions

p \leftrightarrow q \equiv (p \land q) \lor (\sim p \land \sim q)

(p \land q) \lor (\sim p \land q) \lor (\sim q \land r) = q \lor r

[\sim (\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)

(p \Rightarrow q) \Rightarrow [(r \lor p) \Rightarrow (r \lor q)]

[\sim (\sim p \lor q) \lor (p \land r) \land (\sim q \land r)]

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