Q.Without using truth table, prove that: $p \leftrightarrow q \equiv (p \wedge q) \vee (\sim p \wedge \sim q)$

$p\leftrightarrow q \equiv (p\to q)\wedge(q\to p)$ [Biconditional law]

$\equiv (\sim p\vee q)\wedge(\sim q\vee p)$ [Conditional law, applied twice]

$\equiv [(\sim p\vee q)\wedge\sim q]\vee[(\sim p\vee q)\wedge p]$ [Distributive law]

$\equiv [(\sim p\wedge\sim q)\vee(q\wedge\sim q)]\vee[(\sim p\wedge p)\vee(q\wedge p)]$ [Distributive law, applied twice]

$\equiv [(\sim p\wedge\sim q)\vee c]\vee[c\vee(p\wedge q)]$ [Complement law: $q\wedge\sim q\equiv c$, $\sim p\wedge p\equiv c$] …

Students often try to prove this with a truth table despite the instruction to use algebra of statements, or skip naming the laws used at each step, whic …