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

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(A) Step $1$: Simplify the expression inside the square brackets. The expression $(\sim p \land q) \lor (p \land \sim q)$ is the definition of $p \opl

Vedclass · Accepted Answer

$p \land \sim q$

Vedclass · Answer

(A) Step $1$: Simplify the expression inside the square brackets. The expression $(\sim p \land q) \lor (p \land \sim q)$ is the definition of $p \oplus q$ (exclusive $OR$).Step $2$: Substitute this back into the original expression: $(p \lor q) \land (p \oplus q) \land \sim q$.Step $3$: Evaluate the conjunction with $\sim q$. If $\sim q$ is true, then $q$ must be false. Substituting $q = F$ into the expression: $(p \lor F) \land [(\sim p \land F) \lor (p \land \sim F)] \land \sim F$.Step $4$: Simplify: $p \land [F \lor (p \land T)] \land T = p \land p \land T = p$.Step $5$: Re-evaluating the whole expression using a truth table or logical laws: $(p \lor q) \land [(\sim p \land q) \lor (p \land \sim q)] \land \sim q$. Since the term $\sim q$ is present, $q$ must be false. If $q = F$, the expression becomes $(p \lor F) \land [(\sim p \land F) \lor (p \land T)] \land T = p \land [F \lor p] \land T = p \land p = p$. However, checking the options, if $p$ is false, the expression is false. If $p$ is true and $q$ is false, the expression is true. Thus, it is equivalent to $p \land \sim q$.

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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