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If $(p \wedge \sim r) \rightarrow(\sim p \vee q)$ has truth value ' $F$ ', then truth values of $p, q$ and $r$ are respectively
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The correct answer is:
T,F,F
$\because(p \wedge \sim r) \rightarrow(\sim p \vee q)$ has truth value $F$
$\Rightarrow(p \wedge \sim r)$ is $T$ and $(\sim p \vee q)$ is $F$
$\Rightarrow(P$ is $T$ and $\sim r$ is $T)$ and $(\sim p$ is $F$ and $q$ is $F)$
$\Rightarrow P$ is $T, r$ is $F$ and $q$ is $F$
$\Rightarrow(p \wedge \sim r)$ is $T$ and $(\sim p \vee q)$ is $F$
$\Rightarrow(P$ is $T$ and $\sim r$ is $T)$ and $(\sim p$ is $F$ and $q$ is $F)$
$\Rightarrow P$ is $T, r$ is $F$ and $q$ is $F$
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