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

Just as ordinary algebra has laws for manipulating numerical expressions, logic has a fixed list of laws for rewriting statement patterns without redrawing a truth table every time: Idempotent Law (p ∧ p ≡ p, p ∨ p ≡ p), Commutative Law (p ∨ q ≡ q ∨ p, p ∧ q ≡ q ∧ p), Associative Law (grouping of chained ∧'s or ∨'s doesn't matter), Distributive Law (p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) and its ∨-over-∧ mirror), De Morgan's Law, Identity Law (p ∧ t ≡ p, p ∧ c ≡ c, p ∨ c ≡ p, p ∨ t ≡ t), Complement Law (p ∧ ~p ≡ c, p ∨ ~p ≡ t), Absorption Law (p ∨ (p ∧ q) ≡ p, p ∧ (p ∨ q) ≡ p), the Conditional Law (p → q ≡ ~p ∨ q), and the Bicondit …