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

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(A) Given expression: $[\sim(\sim p \vee q) \vee (p \wedge r) \wedge (\sim q \wedge r)]$. Applying De Morgan's Law to the first part: $\sim(\sim p \ve

Vedclass · Accepted Answer

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

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(A) Given expression: $[\sim(\sim p \vee q) \vee (p \wedge r) \wedge (\sim q \wedge r)]$. Applying De Morgan's Law to the first part: $\sim(\sim p \vee q) \equiv (p \wedge \sim q)$. Now the expression becomes: $[(p \wedge \sim q) \vee ((p \wedge r) \wedge (\sim q \wedge r))]$. Using the associative and commutative properties: $(p \wedge r) \wedge (\sim q \wedge r) \equiv (p \wedge \sim q) \wedge (r \wedge r) \equiv (p \wedge \sim q) \wedge r$. Substituting this back: $(p \wedge \sim q) \vee ((p \wedge \sim q) \wedge r)$. Using the absorption law: $A \vee (A \wedge B) \equiv A$,where $A = (p \wedge \sim q)$ and $B = r$. Therefore,the expression simplifies to $(p \wedge \sim q)$.

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

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