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

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Question

(B) To determine the equivalence,we construct the truth table for the expression $(P \vee Q) \wedge (\sim P) \Rightarrow Q$.$P$$Q$$P \vee Q$$\sim P$$(

Vedclass · Accepted Answer

$\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$

Vedclass · Answer

(B) To determine the equivalence,we construct the truth table for the expression $(P \vee Q) \wedge (\sim P) \Rightarrow Q$.$P$$Q$$P \vee Q$$\sim P$$(P \vee Q) \wedge (\sim P)$$(P \vee Q) \wedge (\sim P) \Rightarrow Q$$T$$T$$T$$F$$F$$T$$T$$F$$T$$F$$F$$T$$F$$T$$T$$T$$T$$T$$F$$F$$F$$T$$F$$T$The final column shows that the statement is a tautology (always $T$).Checking the options,option $B$ is $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$. Since $\sim(P \Rightarrow Q) \equiv P \wedge \sim Q$,the biconditional $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$ is also a tautology.Thus,the statement is equivalent to option $B$.

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

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