1. **Problem Statement:** Prove that $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$. 2. **Formula and Important Rules:** - The intersection of sets $$X \cap Y$$ contains elements common to both $$X$$ and $$Y$$. - The union of sets $$X \cup Y$$ contains elements in $$X$$ or $$Y$$ or both. - To prove set equality, show two inclusions: $$A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$$ and $$ (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$$. 3. **Inclusion One: $$A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$$** - Take an arbitrary element $$x \in A \cap (B \cup C)$$. - By definition of intersection, $$x \in A$$ and $$x \in B \cup C$$. - Since $$x \in B \cup C$$, either $$x \in B$$ or $$x \in C$$. - Case 1: If $$x \in B$$, then $$x \in A$$ and $$x \in B$$, so $$x \in A \cap B$$. - Case 2: If $$x \in C$$, then $$x \in A$$ and $$x \in C$$, so $$x \in A \cap C$$. - Therefore, $$x \in (A \cap B) \cup (A \cap C)$$. - Hence, $$A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$$. 4. **Inclusion Two: $$(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$$** - Take an arbitrary element $$x \in (A \cap B) \cup (A \cap C)$$. - By definition of union, $$x \in A \cap B$$ or $$x \in A \cap C$$. - Case 1: If $$x \in A \cap B$$, then $$x \in A$$ and $$x \in B$$. - Case 2: If $$x \in A \cap C$$, then $$x \in A$$ and $$x \in C$$. - In both cases, $$x \in A$$ and $$x \in B \cup C$$ (since $$x$$ is in either $$B$$ or $$C$$). - Therefore, $$x \in A \cap (B \cup C)$$. - Hence, $$(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$$. 5. **Conclusion:** Since both inclusions hold, we conclude that: $$ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $$ Thus, the equality is proven using set inclusion arguments.