Negation of the Boolean statement (p vee q) R | vedclass

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(C) Let the statement be $S = (p \vee q) \Rightarrow ((\sim r) \vee p)$.Using the implication rule $A \Rightarrow B \equiv \sim A \vee B$,we get:$S \e

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

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(C) Let the statement be $S = (p \vee q) \Rightarrow ((\sim r) \vee p)$.Using the implication rule $A \Rightarrow B \equiv \sim A \vee B$,we get:$S \equiv \sim (p \vee q) \vee ((\sim r) \vee p)$$S \equiv (\sim p \wedge \sim q) \vee (\sim r \vee p)$Using the distributive law $(A \wedge B) \vee C \equiv (A \vee C) \wedge (B \vee C)$:$S \equiv (\sim p \vee \sim r \vee p) \wedge (\sim q \vee \sim r \vee p)$Since $(\sim p \vee p) \equiv T$ (Tautology),we have:$S \equiv (T \vee \sim r) \wedge (\sim q \vee \sim r \vee p)$$S \equiv T \wedge (\sim q \vee \sim r \vee p) \equiv \sim q \vee \sim r \vee p$Now,the negation of $S$ is $\sim (\sim q \vee \sim r \vee p)$.Applying De Morgan's Law,$\sim (A \vee B \vee C) \equiv \sim A \wedge \sim B \wedge \sim C$:$\sim S \equiv q \wedge r \wedge \sim p$Rearranging,we get $(\sim p) \wedge q \wedge r$.

Negation of the Boolean statement $(p \vee q) \Rightarrow ((\sim r) \vee p)$ is equivalent to

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