The negation of q vee sim (p wedge r) is | vedclass

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(B) To find the negation of the statement $q \vee \sim (p \wedge r)$,we apply De Morgan's Law: $\sim (A \vee B) = \sim A \wedge \sim B$. Applying this

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$\sim q \wedge (p \wedge r)$

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(B) To find the negation of the statement $q \vee \sim (p \wedge r)$,we apply De Morgan's Law: $\sim (A \vee B) = \sim A \wedge \sim B$. Applying this,we get: $\sim (q \vee \sim (p \wedge r)) = \sim q \wedge \sim (\sim (p \wedge r))$. Using the law of double negation $\sim (\sim P) = P$,we simplify the expression: $\sim q \wedge (p \wedge r)$. Thus,the correct option is $B$.

The negation of $q \vee \sim (p \wedge r)$ is

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