The negation of the statement pattern (p wedg | vedclass

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(C) Let the given statement be $S = (p \wedge \sim q) \rightarrow (p \vee \sim q)$.Recall that the negation of an implication $A \rightarrow B$ is $A

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a contradiction

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(C) Let the given statement be $S = (p \wedge \sim q) \rightarrow (p \vee \sim q)$.Recall that the negation of an implication $A \rightarrow B$ is $A \wedge \sim B$.Here,$A = (p \wedge \sim q)$ and $B = (p \vee \sim q)$.So,$\sim S = (p \wedge \sim q) \wedge \sim (p \vee \sim q)$.Using De Morgan's Law,$\sim (p \vee \sim q) = (\sim p \wedge \sim (\sim q)) = (\sim p \wedge q)$.Thus,$\sim S = (p \wedge \sim q) \wedge (\sim p \wedge q)$.By the associative and commutative properties,$\sim S = (p \wedge \sim p) \wedge (\sim q \wedge q)$.Since $(p \wedge \sim p) = F$ (a contradiction) and $(\sim q \wedge q) = F$,we have $\sim S = F \wedge F = F$.$A$ statement that is always false is called a contradiction.

The negation of the statement pattern $(p \wedge \sim q) \rightarrow (p \vee \sim q)$ is

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