The statement (p wedge q) Rightarrow (p wedge | vedclass

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(D) Given statement: $(p \wedge q) \Rightarrow (p \wedge r)$Using the logical equivalence $A \Rightarrow B \equiv \sim A \vee B$:$\sim(p \wedge q) \ve

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$(p \wedge q) \Rightarrow r$

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(D) Given statement: $(p \wedge q) \Rightarrow (p \wedge r)$Using the logical equivalence $A \Rightarrow B \equiv \sim A \vee B$:$\sim(p \wedge q) \vee (p \wedge r)$Apply De Morgan's Law:$(\sim p \vee \sim q) \vee (p \wedge r)$Using the distributive law $(\sim p \vee \sim q) \vee (p \wedge r) \equiv ((\sim p \vee \sim q) \vee p) \wedge ((\sim p \vee \sim q) \vee r)$:$((\sim p \vee p) \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$Since $(\sim p \vee p) \equiv T$ (Tautology):$(T \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$$T \wedge (\sim p \vee \sim q \vee r) \equiv \sim p \vee \sim q \vee r$Rearranging the terms:$(\sim p \vee \sim q) \vee r$Using De Morgan's Law again:$\sim(p \wedge q) \vee r$Converting back to implication:$(p \wedge q) \Rightarrow r$

The statement $(p \wedge q) \Rightarrow (p \wedge r)$ is equivalent to:

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