Question
Without using truth table prove that:

~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

Answer

We have,
L.H.S. = ∼(p ∨ q) ∨ (∼p ∧ q)
= (∼ p ∧ ∼q) ∨ (∼p ∧ q) ....(By De Morgan's Law)
= ∼p ∧ (∼q ∨ q) ....(By Distributive Law)
= ∼p ∧ T ....(By Complement Law)
= ∼p

R.H.S. = ∼p

L.H.S. = R.H.S.

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