- P - q is logically equivalent toA - - p - q B- -p - qC- P - qD- p - q

The correct option is C p ∨∼ q We know that ∼ (p ∧ q)= ∼ p ∨∼ q ∴ ∼ (∼ p ∧ q) = ∼(∼ p) ∨∼ q =p ∨∼ q ...

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Negation

∼∼ P ∧ q is l...

Question

$\sim (\sim p \land q)$ is logically equivalent to

$\sim (p \lor q)$

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$\sim (p \land \sim q)$

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$p \land \sim q$

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$p \lor \sim q$

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Solution

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Similar questions

\sim (\sim p \Rightarrow q)

\sim (\sim p \to q)

~[(- p)^q] is logically equivalent to

(p \land \sim q) \lor q \lor (\sim p \land q)

(p \land \sim q) \lor q \lor (\sim p \land q)

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