The negation of the contrapositive of the sta | vedclass

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(B) The contrapositive of the statement $(p \vee \sim q) \rightarrow (p \wedge \sim q)$ is given by $\sim (p \wedge \sim q) \rightarrow \sim (p \vee \

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

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(B) The contrapositive of the statement $(p \vee \sim q) \rightarrow (p \wedge \sim q)$ is given by $\sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.Using the equivalence $A \rightarrow B \equiv \sim A \vee B$,we get:$\sim (p \wedge \sim q) \vee \sim (p \vee \sim q)$Applying De Morgan's law:$(\sim p \vee q) \vee (\sim p \wedge q)$Now,we find the negation of this contrapositive:$\sim [(\sim p \vee q) \vee (\sim p \wedge q)]$Using De Morgan's law again:$\sim (\sim p \vee q) \wedge \sim (\sim p \wedge q)$$(p \wedge \sim q) \wedge (p \vee \sim q)$Wait,let us re-evaluate the negation of the contrapositive directly:The contrapositive is $C = \sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.The negation of $A \rightarrow B$ is $A \wedge \sim B$.Thus,$\sim C = \sim (p \wedge \sim q) \wedge \sim [\sim (p \vee \sim q)]$$\sim C = (\sim p \vee q) \wedge (p \vee \sim q)$.

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

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