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 into $S$: $S = (p \lor q) \land (p \oplus q) \land \sim q$.Step $3$: Distribute $\sim q$ into the expression. Since $(p \lor q) \land \sim q$ is equivalent to $(p \land \sim q) \lor (q \land \sim q)$, and $(q \land \sim q)$ is $F$ (False), this simplifies to $(p \land \sim q)$.Step $4$: Now we have $S = (p \land \sim q) \land (p \oplus q)$.Step $5$: Expand $p \oplus q$ as $(\sim p \land q) \lor (p \land \sim q)$. So, $S = (p \land \sim q) \land [(\sim p \land q) \lor (p \land \sim q)]$.Step $6$: By the distributive law, $S = [(p \land \sim q) \land (\sim p \land q)] \lor [(p \land \sim q) \land (p \land \sim q)]$.Step $7$: The first part $(p \land \sim q \land \sim p \land q)$ is $F$ because $(p \land \sim p)$ is $F$. The second part is $(p \land \sim q)$. Thus, $S = F \lor (p \land \sim q) = p \land \sim q$.

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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