The negation of (p wedge sim q) rightarrow (p | vedclass

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

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a contradiction

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(C) Let $S$ be the statement $(p \wedge \sim q) \rightarrow (p \vee \sim q)$.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,the negation is $(p \wedge \sim q) \wedge \sim (p \vee \sim q)$.Using De Morgan's Law,$\sim (p \vee \sim q) \equiv \sim p \wedge \sim (\sim q) \equiv \sim p \wedge q$.Thus,the negation is $(p \wedge \sim q) \wedge (\sim p \wedge q)$.By the associative and commutative properties,this is $(p \wedge \sim p) \wedge (\sim q \wedge q)$.Since $p \wedge \sim p$ is a contradiction $(F)$ and $\sim q \wedge q$ is a contradiction $(F)$,the expression becomes $F \wedge F$,which is $F$.$A$ statement that is always false is called a contradiction.

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

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