Question
Without using truth table show that : $\sim(p \vee q) \vee(\sim p \wedge q) \equiv \sim p$

Answer

~(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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