Using the rules in logic, write the negation of the following: (p &or; q) &and; (q &or; &sim;r)

The negation of (p &or; q) &and; (q &or; &sim;r) is&sim; [(p &or; q) &and; (q &or; &sim;r)]&equiv; &sim;(p &or; q) &or; &sim;(q &or; &sim;r) .....(Negation of conjunction)&equiv; (&sim;p &and; &sim;q) &or; [&sim;q &and; &sim;(&sim;r)] ...............(Negation of disjunction&equiv; (&sim;p &and; &sim;q) &or; (&sim;q &and; r) ...........(Negation of negation)&equiv; (&sim;q &and; &sim;p) &or; (&sim;q &and; r) ..........(Commutative law)&equiv; (&sim;q) &and; (&sim;p &or; r) ..........(Distributive Law)

Using the rules in logic, write the negation of the following: (p ∨ q) ∧ (q ∨ ∼r) - Mathematics and Statistics

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Sum

Using the rules in logic, write the negation of the following:

(p ∨ q) ∧ (q ∨ ∼r)

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Solution

The negation of (p ∨ q) ∧ (q ∨ ∼r) is
∼ [(p ∨ q) ∧ (q ∨ ∼r)]
≡ ∼(p ∨ q) ∨ ∼(q ∨ ∼r) .....(Negation of conjunction)
≡ (∼p ∧ ∼q) ∨ [∼q ∧ ∼(∼r)] ...............(Negation of disjunction
≡ (∼p ∧ ∼q) ∨ (∼q ∧ r) ...........(Negation of negation)
≡ (∼q ∧ ∼p) ∨ (∼q ∧ r) ..........(Commutative law)
≡ (∼q) ∧ (∼p ∨ r) ..........(Distributive Law)

Concept: Algebra of Statements

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Chapter 1: Mathematical Logic - Miscellaneous Exercise 1 [Page 34]

Q 11.1 Q 10.3 Q 11.2