The negation of sim s vee (sim r wedge s) is | vedclass

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(A) We need to find the negation of the expression $\sim s \vee (\sim r \wedge s)$.Let $P = \sim s \vee (\sim r \wedge s)$.The negation is $\sim P = \

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$s \wedge r$

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(A) We need to find the negation of the expression $\sim s \vee (\sim r \wedge s)$.Let $P = \sim s \vee (\sim r \wedge s)$.The negation is $\sim P = \sim (\sim s \vee (\sim r \wedge s))$.Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$:$\sim P \equiv \sim (\sim s) \wedge \sim (\sim r \wedge s)$.Using the double negation law $\sim (\sim s) \equiv s$ and De Morgan's Law $\sim (A \wedge B) \equiv \sim A \vee \sim B$:$\sim P \equiv s \wedge (\sim (\sim r) \vee \sim s)$.Simplifying further:$\sim P \equiv s \wedge (r \vee \sim s)$.Using the distributive law $A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$:$\sim P \equiv (s \wedge r) \vee (s \wedge \sim s)$.Since $(s \wedge \sim s) \equiv F$ (a contradiction):$\sim P \equiv (s \wedge r) \vee F \equiv s \wedge r$.

The negation of $\sim s \vee (\sim r \wedge s)$ is equivalent to

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