1. **State the problem:** Prove that $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$. 2. **Recall the definitions:** - The intersection $$A \cap B$$ is the set of elements that are in both $$A$$ and $$B$$. - The union $$B \cup C$$ is the set of elements that are in $$B$$ or $$C$$ or both. 3. **Show the left side:** An element $$x \in A \cap (B \cup C)$$ means $$x \in A$$ and $$x \in (B \cup C)$$. 4. **Expand the union:** Since $$x \in (B \cup C)$$ means $$x \in B$$ or $$x \in C$$, then $$x \in A$$ and ($$x \in B$$ or $$x \in C$$). 5. **Distribute intersection over union:** This is equivalent to ($$x \in A$$ and $$x \in B$$) or ($$x \in A$$ and $$x \in C$$), which means $$x \in (A \cap B) \cup (A \cap C)$$. 6. **Conclusion:** Since every element of $$A \cap (B \cup C)$$ is in $$(A \cap B) \cup (A \cap C)$$ and vice versa, the two sets are equal. Therefore, $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$.