The negation of the contrapositive of the sta | vedclass

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(D) Let the statement be $S: A 	o B$, where $A = (p \vee \sim q)$ and $B = (p \wedge \sim q)$.The contrapositive of $S$ is $\sim B 	o \sim A$.The ne

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

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(D) Let the statement be $S: A 	o B$, where $A = (p \vee \sim q)$ and $B = (p \wedge \sim q)$.The contrapositive of $S$ is $\sim B 	o \sim A$.The negation of the contrapositive is $\sim (\sim B 	o \sim A)$.Using the identity $\sim (X 	o Y) \equiv X \wedge \sim Y$, we get $\sim B \wedge \sim (\sim A) \equiv \sim B \wedge A$.Substitute $A$ and $B$: $(p \vee \sim q) \wedge \sim (p \wedge \sim q)$.Using De Morgan's law, $\sim (p \wedge \sim q) \equiv \sim p \vee q$.So, the expression is $(p \vee \sim q) \wedge (\sim p \vee q)$.Expanding this: $(p \wedge \sim p) \vee (p \wedge q) \vee (\sim q \wedge \sim p) \vee (\sim q \wedge q)$.Since $(p \wedge \sim p) = F$ and $(\sim q \wedge q) = F$, we get $(p \wedge q) \vee (\sim p \wedge \sim q)$.Wait, re-evaluating the expression $(p \vee \sim q) \wedge (\sim p \vee q)$ directly matches option $(D)$ if we look at the structure, but let's check the logic again. The negation of $A 	o B$ is $A \wedge \sim B$. The contrapositive is $\sim B 	o \sim A$. The negation of the contrapositive is $\sim B \wedge A$. This is equivalent to the negation of the original statement. Thus, the answer is $(p \vee \sim q) \wedge (\sim p \vee q)$.

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

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