The negation of (p land q) to ((p lor r) to s | vedclass

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(A) Let the given statement be $S = (p \land q) 	o ((p \lor r) 	o \sim q)$.Using the implication rule $A 	o B \equiv \sim A \lor B$, we have $S \eq

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$p \land q$

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(A) Let the given statement be $S = (p \land q) 	o ((p \lor r) 	o \sim q)$.Using the implication rule $A 	o B \equiv \sim A \lor B$, we have $S \equiv \sim (p \land q) \lor (\sim (p \lor r) \lor \sim q)$.The negation of $S$ is $\sim S = \sim [\sim (p \land q) \lor (\sim (p \lor r) \lor \sim q)]$.Using De Morgan's law $\sim (A \lor B) = \sim A \land \sim B$, we get $\sim S = (p \land q) \land \sim (\sim (p \lor r) \lor \sim q)$.Applying De Morgan's law again, $\sim S = (p \land q) \land ((p \lor r) \land q)$.Since $(p \land q) \land q = p \land q$, we have $\sim S = (p \land q) \land (p \lor r) = q \land (p \land (p \lor r))$.By the absorption law, $p \land (p \lor r) = p$, so $\sim S = q \land p = p \land q$.

The negation of $(p \land q) \to ((p \lor r) \to \sim q)$ is equivalent to ...

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