Question

Without using truth table, prove that:

[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p

Answer 1

LHS = [p ∧ (q v r)] ∨ [∼r ∧ ∼q ∧ p]

≡ [p ∧ (q ∨ r)] ∨ [(∼r ∧ ∼q) ∧ p]            ... (Associative Law)

≡ [p ∧ (q ∨ r)] ∨ [(∼q ∧ ∼r) ∧ p]          ... (Commutative Law)

≡ [p ∧ (q ∨ r)] ∨ [∼(q ∨ r) ∧ p]        ... (De Morgan’s Law)

≡ [p ∧(q ∨ r)] ∨ [p ∧ ∼(q ∨ r)]           ... (Commutative Law)

≡ p ∧ [(q ∨ r) ∨ ∼(q ∨ r)]          ... (Distributive Law)

≡ p ∧ t              ... (Complement Law)

= RHS