The logical statement [ sim ( sim p vee q) ve | vedclass

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(D) Given expression: $[ \sim ( \sim p \vee q) \vee (p \wedge r) ] \wedge ( \sim q \wedge r)$Using De Morgan's Law,$\sim ( \sim p \vee q) \equiv (p \w

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

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(D) Given expression: $[ \sim ( \sim p \vee q) \vee (p \wedge r) ] \wedge ( \sim q \wedge r)$Using De Morgan's Law,$\sim ( \sim p \vee q) \equiv (p \wedge \sim q)$.So,the expression becomes: $[ (p \wedge \sim q) \vee (p \wedge r) ] \wedge ( \sim q \wedge r)$Using the Distributive Law,$(p \wedge \sim q) \vee (p \wedge r) \equiv p \wedge ( \sim q \vee r)$.Now,the expression is: $[ p \wedge ( \sim q \vee r) ] \wedge ( \sim q \wedge r)$By Associative Law: $p \wedge [ ( \sim q \vee r) \wedge ( \sim q \wedge r) ]$Since $( \sim q \vee r) \wedge ( \sim q \wedge r) \equiv ( \sim q \wedge r)$ (by Absorption Law),the expression simplifies to:$p \wedge ( \sim q \wedge r) \equiv (p \wedge \sim q) \wedge r$.

The logical statement $[ \sim ( \sim p \vee q) \vee (p \wedge r) ] \wedge ( \sim q \wedge r)$ is equivalent to

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