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

We need to determine which statement is equivalent to $$(p \wedge q) \Rightarrow (p \wedge r)$$.

We begin by rewriting the implication using the identity $$A \Rightarrow B \equiv \neg A \vee B$$. We have $$(p \wedge q) \Rightarrow (p \wedge r) \equiv \neg(p \wedge q) \vee (p \wedge r) \equiv (\neg p \vee \neg q) \vee (p \wedge r)$$.

Now let us examine Option D: $$(p \wedge q) \Rightarrow r \equiv \neg(p \wedge q) \vee r \equiv (\neg p \vee \neg q) \vee r$$.

We claim these two expressions are logically equivalent. To verify, we check all cases based on $$p$$.

Case 1: If $$p$$ is false, then $$\neg p$$ is true, so both $$(\neg p \vee \neg q) \vee (p \wedge r)$$ and $$(\neg p \vee \neg q) \vee r$$ evaluate to true regardless of $$q$$ and $$r$$.

Case 2: If $$p$$ is true and $$q$$ is false, then $$\neg q$$ is true, so both expressions again evaluate to true.

Case 3: If $$p$$ is true and $$q$$ is true, then $$\neg p \vee \neg q = \text{false}$$. The first expression becomes $$(p \wedge r) = r$$ (since $$p$$ is true), and the second becomes simply $$r$$. Hence both equal $$r$$.

In every case the two expressions agree, confirming that $$(p \wedge q) \Rightarrow (p \wedge r) \equiv (p \wedge q) \Rightarrow r$$.

Hence, the correct answer is Option D.