We know that $p\leftrightarrow q\equiv (p\rightarrow q)\wedge (q\rightarrow p)$
$p\leftrightarrow q\equiv(\sim p\vee q)\wedge (\sim q\vee p)\equiv(\sim p\wedge\sim q)\vee(p\wedge q)$
$p\leftrightarrow q\equiv(p\wedge q)\vee (\sim p\wedge\sim q)$
lets prove using truth table
$p$ |
$q$ |
$\sim p$ |
$\sim q$ |
$ (p\wedge q)$ |
$(\sim p\wedge\sim q)$ |
$(p\wedge q)\vee(\sim p\wedge\sim q)$ |
$p\leftrightarrow q$ |
T |
T |
F |
F |
T |
F |
T |
T |
T |
F |
F |
T |
F |
F |
F |
F |
F |
T |
T |
F |
F |
F |
F |
F |
F |
F |
T |
T |
F |
T |
T |
T |