The negation of (p Rightarrow q) Rightarrow ( | vedclass

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(C) Let $S = (p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$.Using the identity $A \Rightarrow B \equiv \sim A \vee B$,we have:$S \equiv \sim (

Vedclass · Accepted Answer

$q \wedge (\sim p)$

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(C) Let $S = (p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$.Using the identity $A \Rightarrow B \equiv \sim A \vee B$,we have:$S \equiv \sim (p$ $\Rightarrow q) \vee (q$ $\Rightarrow p)$$S \equiv \sim (\sim p \vee q) \vee (\sim q \vee p)$$S \equiv (p \wedge \sim q) \vee (\sim q \vee p)$Using the associative and commutative laws:$S \equiv (p \vee \sim q \vee p) \wedge (\sim q \vee \sim q \vee p)$$S \equiv (p \vee \sim q) \wedge (\sim q \vee p) \equiv p \vee \sim q$.Now,the negation of $S$ is $\sim (p \vee \sim q)$.By De Morgan's Law,$\sim (p \vee \sim q) \equiv \sim p \wedge \sim (\sim q) \equiv \sim p \wedge q$ or $q \wedge (\sim p)$.

The negation of $(p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$ is

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