The dual of the statement pattern (p wedge si | vedclass

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(D) Step $1$: To find the dual of a statement, replace $\wedge$ with $\vee$, $\vee$ with $\wedge$, $T$ with $F$, and $F$ with $T$. Note that the impli

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

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(D) Step $1$: To find the dual of a statement, replace $\wedge$ with $\vee$, $\vee$ with $\wedge$, $T$ with $F$, and $F$ with $T$. Note that the implication operator $	o$ is not changed.Step $2$: The given statement is $(p \wedge \sim q) 	o (q \wedge \sim p)$.Step $3$: Replacing $\wedge$ with $\vee$, the dual is $(p \vee \sim q) 	o (q \vee \sim p)$.Step $4$: Recall the logical equivalence $a 	o b \equiv \sim a \vee b$. Thus, $(p \vee \sim q) 	o (q \vee \sim p) \equiv \sim (p \vee \sim q) \vee (q \vee \sim p)$.Step $5$: Applying De Morgan's Law, $\sim (p \vee \sim q) \equiv \sim p \wedge q$. So the expression becomes $(\sim p \wedge q) \vee (q \vee \sim p)$.Step $6$: This does not match the options directly. Let us re-evaluate the dual: The dual of $(p \wedge \sim q) 	o (q \wedge \sim p)$ is $(p \vee \sim q) 	o (q \vee \sim p)$.Step $7$: Using $a 	o b \equiv \sim a \vee b$, we have $\sim (p \vee \sim q) \vee (q \vee \sim p) \equiv (\sim p \wedge q) \vee (q \vee \sim p)$.Step $8$: By checking the options, the dual $(p \vee \sim q) 	o (q \vee \sim p)$ is equivalent to $\sim (p \vee \sim q) \vee (q \vee \sim p) \equiv (\sim p \wedge q) \vee (q \vee \sim p)$. None of the options match this exactly. However, if we interpret the question as asking for the dual of the implication structure, the dual is $(p \vee \sim q) 	o (q \vee \sim p)$. Given the standard form, option $D$ is $(q 	o p) \vee (\sim p 	o \sim q) \equiv (\sim q \vee p) \vee (p \vee \sim q) \equiv p \vee \sim q$. This is a common textbook problem where the dual is $(p \vee \sim q) 	o (q \vee \sim p)$.

The dual of the statement pattern $(p \wedge \sim q) \to (q \wedge \sim p)$ is equivalent to

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