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Question
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
Solution
~(p v q)v(~p ∧ q)
≡~(p v q)v~(p ∨ ~q) by De Morgan's Law
≡~[(p ∨ q) ∧ (p ∨ ~q)] by De Morgan's Law
≡~{[(p ∨ q) ∧ p] ∨ [(p ∨ q)∧ ~q)]} by Distributive Law
≡ ~{[p] ∨ [(p ∨ q) ∧ ~q]} by Absorption Law
≡ ~{[p] ∨ [(p∧ ~q) ∨ (q ∧ ~q)]} by Distributive Law
≡~{[p] ∨ [(p ∧ ~q) ∨ F]} by Complement Law
≡~{[p] ∨ [(p ∧ ~q)]} by Identity Law
≡~p ∧ (~p ∨ q) by De Morgan's Law
≡ ~p by Absorption Law
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