If the truth value of the expression [(p vee | vedclass

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(B) The implication $A \rightarrow B$ is False if and only if $A$ is True and $B$ is False.Here,$A = [(p \vee q) \wedge (q \rightarrow r) \wedge (\sim

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$T, F, F$

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(B) The implication $A \rightarrow B$ is False if and only if $A$ is True and $B$ is False.Here,$A = [(p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)]$ and $B = (p \wedge q)$.For $B = (p \wedge q)$ to be False,at least one of $p$ or $q$ must be False.For $A$ to be True,all components $(p \vee q)$,$(q \rightarrow r)$,and $(\sim r)$ must be True.From $(\sim r) = T$,we get $r = F$.Substituting $r = F$ into $(q \rightarrow r) = T$,we get $(q \rightarrow F) = T$,which implies $q = F$.Now,substitute $q = F$ into $(p \vee q) = T$,we get $(p \vee F) = T$,which implies $p = T$.Checking $B = (p \wedge q) = (T \wedge F) = F$,which satisfies the condition.Thus,the truth values are $p = T, q = F, r = F$.

If the truth value of the expression $[(p \vee q) \wedge (q$ $\rightarrow r) \wedge (\sim r)]$ $\rightarrow (p \wedge q)$ is False,then the truth values of $p, q, r$ are respectively:

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