The Boolean expression ((p wedge q) vee (p ve | vedclass

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(C) Let the expression be $E = ((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$.Using the associative and commutative laws,we simpli

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

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(C) Let the expression be $E = ((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$.Using the associative and commutative laws,we simplify the first part: $(p \wedge q) \vee (p \vee \sim q) \equiv (p \vee (p \wedge q)) \vee \sim q \equiv p \vee \sim q$.Now,substitute this back into the expression: $E \equiv (p \vee \sim q) \wedge (\sim p \wedge \sim q)$.Using the distributive law: $E \equiv (p \wedge (\sim p \wedge \sim q)) \vee (\sim q \wedge (\sim p \wedge \sim q))$.Since $p \wedge \sim p \equiv F$ (False),the first term becomes $F \wedge \sim q \equiv F$.The second term simplifies as $\sim q \wedge \sim q \equiv \sim q$,so we have $F \vee (\sim p \wedge \sim q)$.Thus,$E \equiv \sim p \wedge \sim q$.

The Boolean expression $((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$ is equivalent to

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