If (p wedge sim r) rightarrow (sim p vee q) h | vedclass

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Question

(C) The implication $A \rightarrow B$ is $False$ only when $A$ is $True$ and $B$ is $False$.Given $(p \wedge \sim r) \rightarrow (\sim p \vee q) \equi

Vedclass · Accepted Answer

$T, F, F$

Vedclass · Answer

(C) The implication $A \rightarrow B$ is $False$ only when $A$ is $True$ and $B$ is $False$.Given $(p \wedge \sim r) \rightarrow (\sim p \vee q) \equiv F$.This implies $(p \wedge \sim r) \equiv T$ and $(\sim p \vee q) \equiv F$.From $(\sim p \vee q) \equiv F$,we get $\sim p \equiv F$ and $q \equiv F$,which means $p \equiv T$ and $q \equiv F$.Substituting $p \equiv T$ into $(p \wedge \sim r) \equiv T$,we get $(T \wedge \sim r) \equiv T$,which implies $\sim r \equiv T$,so $r \equiv F$.Thus,the truth values are $p=T, q=F, r=F$.Therefore,Option $(C)$ is correct.

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

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