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Question
Without using truth table, show that
p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p
Solution
L.H.S.
≡ p ∧ [(~ p ∨ q) ∨ ~ q]
≡ p ∧ [(~ p ∨ (q ∨ ~ q)] .....[Associative law]
≡ p ∧ (~ p ∨ T) .....[Complement law]
≡ p ∧ T .....[Identity law]
≡ p .....[Identity law]
≡ R.H.S.
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