The logical statement (p wedge sim q) vee q v | vedclass

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(D) Let the given statement be $S = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.Using the associative and commutative laws,we can rearrange the t

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$p \vee q$

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(D) Let the given statement be $S = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.Using the associative and commutative laws,we can rearrange the terms:$S = \{(p \wedge \sim q) \vee (\sim p \wedge q)\} \vee q$We know that $(p \wedge \sim q) \vee (\sim p \wedge q)$ is the logical expression for the exclusive $OR$,denoted as $p \oplus q$ or $\sim(p \Leftrightarrow q)$.So,$S = (p \oplus q) \vee q$.Using the distributive law: $(p \oplus q) \vee q \equiv (p \vee q) \wedge (\sim q \vee q) \equiv (p \vee q) \wedge T \equiv p \vee q$.Alternatively,using a truth table:| $p$ | $q$ | $p \wedge \sim q$ | $\sim p \wedge q$ | $(p \wedge \sim q) \vee (\sim p \wedge q)$ | $S = \{(p \wedge \sim q) \vee (\sim p \wedge q)\} \vee q$ | $p \vee q$ ||---|---|---|---|---|---|---|| $T$ | $T$ | $F$ | $F$ | $F$ | $T$ | $T$ || $T$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ || $F$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ || $F$ | $F$ | $F$ | $F$ | $F$ | $F$ | $F$ |Since the columns for $S$ and $p \vee q$ are identical,the statement is equivalent to $p \vee q$.

The logical statement $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ is equivalent to

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