Question

Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.

Answer 1

≅ [(p ∧ ∼p) ∨ (q ∧ ∼p)] → q     ...(Distributive Law)

≅ [c ∨ (q ∧ ∼p)]→ q                ...(Complement Law)

≅ (q ∧ ∼p) → q               ...(Identity Law)

≅ ( ∼p ∧ q) → q               ...(Commutative Law)

≅ ∼( ∼p ∧ q ) ∨ q          ... (p→q ≅ ∼p ∨ q)

≅ ( p ∨ ∼q) ∨ q                 ...(De Morgan’s Law)

≅ p ∨ ( ∼q ∨ q)                ...(Associative Law)

≅ p ∨ t                 ...(Complement Law)

≅ t           ...(Identity Law)

Hence, [(p ∨ q) ∧ ∼p] →q is a tautology.