1. **Problem statement:** Prove the Distributive Law: $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$. 2. **Formula and rules:** The distributive law in set theory states that intersection distributes over union. This means the intersection of a set with the union of two sets equals the union of the intersections. 3. **Step-by-step proof:** - Start with the left-hand side (LHS): $$A \cap (B \cup C)$$. - By definition of union, $$x \in B \cup C$$ means $$x \in B$$ or $$x \in C$$. - So, $$x \in A \cap (B \cup C)$$ means $$x \in A$$ and ($$x \in B$$ or $$x \in C$$). - Using distributive property of logical AND over OR, this is equivalent to: ($$x \in A$$ and $$x \in B$$) or ($$x \in A$$ and $$x \in C$$). - By definition of intersection, this is: $$x \in (A \cap B) \cup (A \cap C)$$. 4. **Conclusion:** Since every element $$x$$ in the LHS is in the RHS and vice versa, the sets are equal: $$ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $$ This completes the proof of the distributive law for sets.