If the truth value of the compound statement | vedclass

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

Vedclass · Accepted Answer

$(F, T)$

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(C) The implication $A 	o B$ is $False$ only when $A$ is $True$ and $B$ is $False$.Here, $A = [(p \vee q) \wedge (q 	o r) \wedge (\sim r)]$ and $B = (p \wedge q)$.Since $B$ is $False$, $(p \wedge q)$ is $False$.Since $A$ is $True$, all components $(p \vee q)$, $(q 	o r)$, and $(\sim r)$ must be $True$.From $(\sim r) = True$, we get $r = False$.Substitute $r = False$ into $(q 	o r) = True$, which becomes $(q 	o False) = True$. This implies $q = False$.Substitute $q = False$ into $(p \vee q) = True$, which becomes $(p \vee False) = True$. This implies $p = True$.Now, evaluate the required truth values:$p 	o q$ becomes $True 	o False$, which is $False$.$q 	o p$ becomes $False 	o True$, which is $True$.Thus, the truth values are $(F, T)$.

If the truth value of the compound statement $[(p \vee q) \wedge (q \to r) \wedge (\sim r)] \to (p \wedge q)$ is $False$, then the truth values of $p \to q$ and $q \to p$ are respectively......

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