The compound statement (sim(P wedge Q)) vee ( | vedclass

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(B) Let the given statement be $r \Rightarrow s$,where $r = (\sim(P \wedge Q)) \vee ((\sim P) \wedge Q)$ and $s = ((\sim P) \wedge (\sim Q))$.We const

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$(\sim Q) \vee P$

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(B) Let the given statement be $r \Rightarrow s$,where $r = (\sim(P \wedge Q)) \vee ((\sim P) \wedge Q)$ and $s = ((\sim P) \wedge (\sim Q))$.We construct the truth table for the statement:| $P$ | $Q$ | $\sim(P \wedge Q)$ | $(\sim P) \wedge Q$ | $r$ | $s$ | $r \Rightarrow s$ ||---|---|---|---|---|---|---|| $T$ | $T$ | $F$ | $F$ | $F$ | $F$ | $T$ || $T$ | $F$ | $T$ | $F$ | $T$ | $F$ | $F$ || $F$ | $T$ | $T$ | $T$ | $T$ | $F$ | $F$ || $F$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ |Comparing the truth values of $r \Rightarrow s$ with the options:For option $(B)$,$(\sim Q) \vee P$:- If $P=T, Q=T$,$(\sim T) \vee T = F \vee T = T$.- If $P=T, Q=F$,$(\sim F) \vee T = T \vee T = T$.- If $P=F, Q=T$,$(\sim T) \vee F = F \vee F = F$.- If $P=F, Q=F$,$(\sim F) \vee F = T \vee F = T$.This does not match. Let's re-evaluate $r \Rightarrow s$:$r = (\sim P \vee \sim Q) \vee (\sim P \wedge Q) = \sim P \vee (\sim Q \vee (\sim P \wedge Q)) = \sim P \vee (\sim Q \vee \sim P) \wedge (\sim Q \vee Q) = \sim P \vee \sim Q = \sim(P \wedge Q)$.$s = \sim P \wedge \sim Q = \sim(P \vee Q)$.So,$r \Rightarrow s$ is $\sim(P \wedge Q) \Rightarrow \sim(P \vee Q)$.This is equivalent to $\sim(\sim(P \vee Q)) \Rightarrow \sim(\sim(P \wedge Q))$,which is $(P \vee Q) \Rightarrow (P \wedge Q)$.This is only true when $P$ and $Q$ have the same truth value,i.e.,$P \Leftrightarrow Q$.

The compound statement $(\sim(P \wedge Q)) \vee ((\sim P) \wedge Q) \Rightarrow ((\sim P) \wedge (\sim Q))$ is equivalent to

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