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

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Question

(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 get $S \equ

Vedclass · Accepted Answer

$p \land q$

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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 get $S \equiv \sim(p \land q) \lor ((p \lor r) 	o \sim q)$.Applying the rule again to the second part: $S \equiv \sim(p \land q) \lor (\sim(p \lor r) \lor \sim q)$.The negation of $S$ is $\sim S \equiv \sim(\sim(p \land q) \lor (\sim(p \lor r) \lor \sim q))$.By De Morgan's Law, $\sim S \equiv (p \land q) \land \sim(\sim(p \lor r) \lor \sim q)$.$\sim S \equiv (p \land q) \land ((p \lor r) \land q)$.Since $(p \land q) \land q \equiv p \land q$, we have $\sim S \equiv (p \land q) \land (p \lor r)$.Since $(p \land q) \implies p$, and $p \implies (p \lor r)$, the expression simplifies to $p \land q$.

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

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