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(C) Given that the truth value of the implication $(p \wedge \sim r) \rightarrow (\sim p \vee q)$ is $F$. An implication $A \rightarrow B$ is $F$ only

Vedclass · Accepted Answer

$T, F, F$

Vedclass · Answer

(C) Given that the truth value of the implication $(p \wedge \sim r) \rightarrow (\sim p \vee q)$ is $F$. An implication $A \rightarrow B$ is $F$ only when $A$ is $T$ and $B$ is $F$. Therefore,$(p \wedge \sim r)$ is $T$ and $(\sim p \vee q)$ is $F$. For $(p \wedge \sim r)$ to be $T$,both $p$ must be $T$ and $\sim r$ must be $T$. If $p$ is $T$,then $\sim p$ is $F$. For $(\sim p \vee q)$ to be $F$ given $\sim p$ is $F$,$q$ must also be $F$. Since $\sim r$ is $T$,$r$ must be $F$. Thus,the truth values are $p = T, q = F, r = F$.

If $(p \wedge \sim r) \rightarrow (\sim p \vee q)$ has truth value $F$,then the truth values of $p, q,$ and $r$ are respectively:

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If $(p \wedge \sim r) \rightarrow (\sim p \vee q)$ has truth value $F$,then the truth values of $p, q,$ and $r$ are respectively:

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