1. **State the problem:** Prove that $$A \cap (B - C) = (A \cap B) - (A \cap C)$$ where $$B - C = \{x \mid x \in B \text{ and } x \notin C\}$$. 2. **Rewrite the left-hand side (LHS):** $$A \cap (B - C) = \{x \mid x \in A \text{ and } x \in B - C\} = \{x \mid x \in A, x \in B, \text{ and } x \notin C\}$$. 3. **Rewrite the right-hand side (RHS):** $$(A \cap B) - (A \cap C) = \{x \mid x \in A \cap B \text{ and } x \notin A \cap C\} = \{x \mid x \in A, x \in B, \text{ and } x \notin A \cap C\}$$. 4. **Simplify the condition $$x \notin A \cap C$$:** Since $$x \in A$$, $$x \notin A \cap C$$ means $$x \notin C$$. 5. **Therefore, RHS becomes:** $$\{x \mid x \in A, x \in B, \text{ and } x \notin C\}$$. 6. **Compare LHS and RHS:** Both sets are $$\{x \mid x \in A, x \in B, \text{ and } x \notin C\}$$. 7. **Conclusion:** Since LHS and RHS describe the same set, we have proved: $$A \cap (B - C) = (A \cap B) - (A \cap C)$$.