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Distributive Law

Without using...

Question

Without using truth tables, show that
$(p \land q) \lor (\sim p \land q) \lor (\sim q \land r) = q \lor r$

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Solution

$[(p \land q) \lor (\sim p \land q)] \lor (\sim q \land r)$
$= [q \land (p \lor \sim p)] \lor (\sim q \land r)$
$= q \lor (\sim q) \land r$
$= (q \lor \sim q) \land (q \lor r)$
$= q \lor r$ proved

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