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If $p, q, r$ are single propositions with truth values $T, F, F, \quad$ then the truth value of $(p \wedge \sim q) \rightarrow(\sim p \vee r)$ is
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The correct answer is:
$\bar{F}$
\begin{array}{c|c|c|c|c|c|c|c}
\hlinep & q & r & \sim q & \sim p & p \wedge \sim q & \sim p \vee r & (p \wedge \sim q) \rightarrow \\
& & & & & & & (\sim p \vee r) \\
\hlineT & F & F & T & F & T & F & F \\
\hline
\end{array}
\hlinep & q & r & \sim q & \sim p & p \wedge \sim q & \sim p \vee r & (p \wedge \sim q) \rightarrow \\
& & & & & & & (\sim p \vee r) \\
\hlineT & F & F & T & F & T & F & F \\
\hline
\end{array}
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