If the truth value of the Boolean expression | vedclass

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(C) 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

Vedclass · Accepted Answer

$T, F, F$

Vedclass · Answer

(C) 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. Since $(\sim r)$ is true,$r$ must be false. Since $(q \rightarrow r)$ is true and $r$ is false,$q$ must be false. Since $(p \vee q)$ is true and $q$ is false,$p$ must be true. Thus,the truth values are $p=T, q=F, r=F$. Checking option $C$: $p=T, q=F, r=F$ gives $A = (T \vee F) \wedge (F \rightarrow F) \wedge (\sim F) = T \wedge T \wedge T = T$ and $B = (T \wedge F) = F$. Since $T \rightarrow F$ is false,option $C$ is correct.

If the truth value of the Boolean expression $((p \vee q) \wedge (q$ $\rightarrow r) \wedge (\sim r))$ $\rightarrow (p \wedge q)$ is false,then the truth values of the statements $p, q, r$ respectively can be:

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