The statement (p wedge sim q) wedge (sim p ve | vedclass

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(A) Let $S = (p \wedge \sim q) \wedge (\sim p \vee q)$.Using the distributive law,we expand the expression:$S = [(p \wedge \sim q) \wedge \sim p] \vee

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Contradiction

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(A) Let $S = (p \wedge \sim q) \wedge (\sim p \vee q)$.Using the distributive law,we expand the expression:$S = [(p \wedge \sim q) \wedge \sim p] \vee [(p \wedge \sim q) \wedge q]$Using the associative and commutative laws:$S = [(p \wedge \sim p) \wedge \sim q] \vee [p \wedge (\sim q \wedge q)]$Since $(p \wedge \sim p) = c$ (contradiction) and $(\sim q \wedge q) = c$:$S = (c \wedge \sim q) \vee (p \wedge c)$$S = c \vee c = c$Since the result is a contradiction,the statement is a contradiction.

The statement $(p \wedge \sim q) \wedge (\sim p \vee q)$ is a...

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