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

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

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$p \land q$

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(A) Let $S = (p \land q) 	o ((p \lor r) 	o \sim q)$.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 $\sim S \equiv \sim [\sim (p \land q) \lor (\sim (p \lor r) \lor \sim q)]$.Applying De Morgan's Law: $\sim S \equiv (p \land q) \land \sim (\sim (p \lor r) \lor \sim q)$.Applying De Morgan's Law again: $\sim S \equiv (p \land q) \land ((p \lor r) \land q)$.Since $(p \land q) \land q \equiv p \land q$, we get $\sim S \equiv (p \land q) \land (p \lor r)$.Since $(p \land q) \implies (p \lor r)$ is a tautology, $(p \land q) \land (p \lor r) \equiv p \land q$.

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

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