Question
Without using truth table, show that : p ∧ [~p ∨ q) ∨ (~q)] ≡ p

Answer

LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.

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