The logical statement (p lor q) land [(sim p | vedclass

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(A) Let the given expression be $S = (p \lor q) \land [(\sim p \land q) \lor (p \land \sim q)] \land \sim q$.Step $1$: Simplify the middle bracket. Th

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$p \land \sim q$

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(A) Let the given expression be $S = (p \lor q) \land [(\sim p \land q) \lor (p \land \sim q)] \land \sim q$.Step $1$: Simplify the middle bracket. The expression $(\sim p \land q) \lor (p \land \sim q)$ is the definition of the exclusive $OR$ operation, $p \oplus q$.Step $2$: Substitute this back into the expression: $S = (p \lor q) \land (p \oplus q) \land \sim q$.Step $3$: Use the distributive property or truth table. If $\sim q$ is true, then $q$ is false. Substituting $q = F$ into the expression:$S = (p \lor F) \land (p \oplus F) \land T$$S = p \land p \land T = p$.However, checking the options, let's evaluate the truth table for $S$:If $p=T, q=F$: $S = (T \lor F) \land [(F \land F) \lor (T \land T)] \land T = T \land T \land T = T$.If $p=F, q=F$: $S = (F \lor F) \land [(T \land F) \lor (F \land T)] \land T = F \land F \land T = F$.Since $S$ is true only when $p$ is true and $q$ is false, $S \equiv p \land \sim q$.

The logical statement $(p \lor q) \land [(\sim p \land q) \lor (p \land \sim q)] \land \sim q$ is logically equivalent to ...

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