1. The problem is to prove the set identity: $$A \cap (B - C) = (A \cap B) - (A \cap C)$$. 2. Recall that the set difference $B - C$ is defined as $\{x \mid x \in B \text{ and } x \notin C\}$. 3. Therefore, the left side $A \cap (B - C)$ is the set of elements $x$ such that $x \in A$ and $x \in B - C$. 4. By the definition of $B - C$, this means $x \in A$, $x \in B$, and $x \notin C$. 5. Now consider the right side $(A \cap B) - (A \cap C)$. 6. The set $A \cap B$ contains elements $x$ such that $x \in A$ and $x \in B$. 7. The set $A \cap C$ contains elements $x$ such that $x \in A$ and $x \in C$. 8. The difference $(A \cap B) - (A \cap C)$ contains elements $x$ such that $x \in A \cap B$ and $x \notin A \cap C$. 9. This means $x \in A$, $x \in B$, and $x \notin C$ (since if $x$ were in $A \cap C$, it would be in $C$). 10. Both sides describe the set of elements $x$ such that $x \in A$, $x \in B$, and $x \notin C$. 11. Hence, $$A \cap (B - C) = (A \cap B) - (A \cap C)$$ is proven. Final answer: The identity holds true.