Negation of p wedge (q wedge sim(p wedge q)) | vedclass

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(D) We need to find the negation of the statement $S = p \wedge (q \wedge \sim(p \wedge q))$.Using De Morgan's Law,$\sim(A \wedge B) = \sim A \vee \si

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$(\sim(p \wedge q)) \vee p$

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(D) We need to find the negation of the statement $S = p \wedge (q \wedge \sim(p \wedge q))$.Using De Morgan's Law,$\sim(A \wedge B) = \sim A \vee \sim B$ and $\sim(A \vee B) = \sim A \wedge \sim B$.$\sim S = \sim [p \wedge (q \wedge \sim(p \wedge q))]$$= \sim p \vee \sim (q \wedge \sim(p \wedge q))$$= \sim p \vee (\sim q \vee \sim(\sim(p \wedge q)))$$= \sim p \vee (\sim q \vee (p \wedge q))$Using the distributive law,$\sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q)$.Since $(\sim q \vee q) = T$ (Tautology),the expression becomes $(\sim q \vee p) \wedge T = \sim q \vee p$.Thus,$\sim S = \sim p \vee (\sim q \vee p)$.Rearranging the terms,we get $(\sim p \vee p) \vee \sim q = T \vee \sim q = T$.Wait,let us re-evaluate the original expression $p \wedge (q \wedge \sim(p \wedge q))$.$p \wedge (q \wedge (\sim p \vee \sim q)) = p \wedge ((q \wedge \sim p) \vee (q \wedge \sim q)) = p \wedge ((q \wedge \sim p) \vee F) = p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$.The negation of a contradiction $F$ is a tautology $T$. However,looking at the options,let us re-check the simplification.$\sim [p \wedge (q \wedge (\sim p \vee \sim q))] = \sim [ (p \wedge q) \wedge (\sim p \vee \sim q) ] = \sim [ (p \wedge q \wedge \sim p) \vee (p \wedge q \wedge \sim q) ] = \sim [ F \vee F ] = \sim F = T$.Given the options provided,there might be a typo in the question or options. If the question intended to ask for the negation of $p \wedge q$,the answer would be $\sim p \vee \sim q$. Given the structure,option $D$ is the closest logical form.

Negation of $p \wedge (q \wedge \sim(p \wedge q))$ is

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