Share

# Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p - Mathematics and Statistics

ConceptMathematical Logic Algebra of Statements

#### 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

Is there an error in this question or solution?

#### APPEARS IN

2015-2016 (March) (with solutions)
Question 3.1.2 | 3.00 marks
Solution Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p Concept: Mathematical Logic - Algebra of Statements.
S