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, $A = (p \lor q) \land (q 	o r) \land (\sim r)$ and $B = (p

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$(F, T)$

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(C) The implication $A 	o B$ is $False$ only when $A$ is $True$ and $B$ is $False$.Here, $A = (p \lor q) \land (q 	o r) \land (\sim r)$ and $B = (p \land q)$.For $A$ to be $True$, all components must be $True$: $(p \lor q) = T$, $(q 	o r) = T$, and $(\sim r) = T$.From $(\sim r) = T$, we get $r = F$.Since $(q 	o r) = T$ and $r = F$, $q$ must be $F$ (because $T 	o F$ is $F$).Since $(p \lor q) = T$ and $q = F$, $p$ must be $T$.Now, check $B = (p \land q) = (T \land F) = F$. This matches the condition.Thus, $p = T$ and $q = F$.Now, $p 	o q = T 	o F = F$ and $q 	o p = F 	o T = T$.The truth values are $(F, T)$.

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

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