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

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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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Chapter 1.1: Mathematical Logic - Short Answers I

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