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Question
Without using truth table prove that:
~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p
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Solution
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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