#### Question

Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p

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#### 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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Solution for question: Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p concept: null - Algebra of Statements. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Arts, HSC Science (General)