Question

Using the truth table, prove the following logical equivalence:
$p\leftrightarrow q\equiv (p\wedge q)\vee (\sim p\wedge \sim q)$.

Answer 1

$p$	$q$	$p\leftrightarrow q$	$p\wedge q$	$\sim p$	$\sim q$	$\sim p\wedge \sim q$	$(p\wedge q)\vee (\sim p\wedge \sim q)$
$T$	$T$	$T$	$T$	$F$	$F$	$F$	$T$
$T$	$F$	$F$	$F$	$F$	$T$	$F$	$F$
$F$	$T$	$F$	$F$	$T$	$F$	$F$	$F$
$F$	$F$	$T$	$F$	$T$	$T$	$T$	$T$

From column $3$ and $8$ we get $p\leftrightarrow q\equiv (p\wedge q)\vee (\sim p\wedge \sim q)$

Hence proved.