DATA:¬ ≡ NOT ≡ ̅ OR ≡ ∨ ≡ +AND ≡ ∧ ≡ .Formula:a → b ≡ a̅ + bCalculation:S1 : (¬P ∧ (p ∨ q)) → q S1 ≡ (p̅.(p + q) → qS1 ≡ (p̅.p + p̅.q) → qS1 ≡ (p̅.q) → qS1 ≡ \(\overline{\overline p.q} + q\)S1 ≡ p + q̅ + q ≡ p + 1 ≡ 1S1 is a tautologyS2 ≡ q → (¬p ∧ (p ∨ q))S2 ≡ q → (p̅.(p + q))S2 ≡ q̅ + (p̅.p + p̅.q)) S2 ≡ q̅ + p̅.q ≡ p̅ + q̅ S2 is not a tautology.Therefore option 4 is correct