(p wedge sim q) wedge (sim p wedge q) is | vedclass

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(B) Given expression: $(p \wedge \sim q) \wedge (\sim p \wedge q)$By using the associative and commutative laws of logic,we can rearrange the terms:$=

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$A$ contradiction

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(B) Given expression: $(p \wedge \sim q) \wedge (\sim p \wedge q)$By using the associative and commutative laws of logic,we can rearrange the terms:$= (p \wedge \sim p) \wedge (\sim q \wedge q)$We know that for any proposition $p$,$(p \wedge \sim p) = F$ (a contradiction).$= F \wedge F = F$Since the final truth value is always $F$ (False),the statement is a contradiction.

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

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