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 get:$S \equiv \sim(p \land q) \lor

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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 get:$S \equiv \sim(p \land q) \lor (\sim(p \lor r) \lor \sim q)$.$S \equiv (\sim p \lor \sim q) \lor ((\sim p \land \sim r) \lor \sim q)$.By associative and commutative laws, $S \equiv \sim p \lor \sim q \lor (\sim p \land \sim r)$.Since $(\sim p \land \sim r) \implies \sim p$, the expression simplifies to $S \equiv \sim p \lor \sim q$.The negation of $S$ is $\sim(\sim p \lor \sim q)$.By De Morgan's law, $\sim(\sim p \lor \sim q) \equiv p \land q$.

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

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