If (p wedge sim q) wedge r to sim r is F,then | vedclass

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(A) The given logical statement is $(p \wedge \sim q) \wedge r 	o \sim r = F$.An implication $A 	o B$ is false $(F)$ only when $A$ is true $(T)$ and

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$T$

Vedclass · Answer

(A) The given logical statement is $(p \wedge \sim q) \wedge r 	o \sim r = F$.An implication $A 	o B$ is false $(F)$ only when $A$ is true $(T)$ and $B$ is false $(F)$.Therefore,$(p \wedge \sim q) \wedge r = T$ and $\sim r = F$.From $\sim r = F$,we get $r = T$.Checking the first part: if $r = T$,then $(p \wedge \sim q) \wedge T = T$,which implies $(p \wedge \sim q) = T$.This is possible if $p = T$ and $q = F$.Since the condition holds for $r = T$,the truth value of $r$ is $T$.

If $(p \wedge \sim q) \wedge r \to \sim r$ is $F$,then the truth value of $r$ is:

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