Ques. -9

Logic Level 2

One possible expression of the negation of the compound proposition p ∨ ( ∼ p ∨ q ) p\vee (\sim p \vee q) is :

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( p ∧ ∼ q ) ∨ ∼ p (p \wedge \sim q) \vee \sim p None of the given propositions ( p ∨ ∼ q ) ∨ ∼ p (p \vee \sim q) \vee \sim p ( p ∧ ∼ q ) ∧ ∼ p (p \wedge \sim q) \wedge \sim p

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Karan Shekhawat
Feb 25, 2015

Use De-Morgan's Theoram .

~(A AND B)=(~A OR ~B)

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Pranjal Jain
Feb 25, 2015

∼ ( p ∨ ( ∼ p ∨ q ) ) = ∼ p ∧ ∼ ( ∼ p ∨ q ) = ∼ p ∧ ( p ∧ ∼ q ) \sim(p\vee(\sim p\vee q))\\=\sim p\wedge\sim(\sim p\vee q)\\=\sim p\wedge (p\wedge\sim q)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Then , Is this proposition always False ??

Sivaramakrishnan Sivakumar - 6 years, 2 months ago

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