The negation of the statement q wedge (sim p | vedclass

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(A) To find the negation of the statement $q \wedge (\sim p \vee \sim r)$,we apply De Morgan's laws and the negation rules for logical connectives.Let

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

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(A) To find the negation of the statement $q \wedge (\sim p \vee \sim r)$,we apply De Morgan's laws and the negation rules for logical connectives.Let the statement be $S = q \wedge (\sim p \vee \sim r)$.The negation is $\sim S = \sim (q \wedge (\sim p \vee \sim r))$.Using De Morgan's law $\sim (A \wedge B) = \sim A \vee \sim B$,we get:$\sim S = \sim q \vee \sim (\sim p \vee \sim r)$.Using De Morgan's law $\sim (A \vee B) = \sim A \wedge \sim B$,we get:$\sim S = \sim q \vee (\sim (\sim p) \wedge \sim (\sim r))$.Since $\sim (\sim p) = p$ and $\sim (\sim r) = r$,the expression simplifies to:$\sim S = \sim q \vee (p \wedge r)$.Thus,the correct option is $A$.

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

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