1. **State the problem:** Prove that $ (A \cup B) - (A \cap B) = (A - B) \cup (B - A) $. 2. **Recall definitions:** - Union: $ A \cup B = \{x \mid x \in A \text{ or } x \in B\} $ - Intersection: $ A \cap B = \{x \mid x \in A \text{ and } x \in B\} $ - Set difference: $ A - B = \{x \mid x \in A \text{ and } x \notin B\} $ 3. **Rewrite left side:** $$ (A \cup B) - (A \cap B) = \{x \mid x \in A \cup B \text{ and } x \notin A \cap B\} $$ 4. **Express conditions:** - $x \in A \cup B$ means $x \in A$ or $x \in B$ - $x \notin A \cap B$ means $x \notin A$ or $x \notin B$ 5. **Combine conditions:** $$ x \in (A \cup B) - (A \cap B) \iff (x \in A \text{ or } x \in B) \text{ and } (x \notin A \text{ or } x \notin B) $$ 6. **Distribute and simplify:** This is equivalent to: $$ (x \in A \text{ and } x \notin B) \text{ or } (x \in B \text{ and } x \notin A) $$ 7. **Recognize right side:** - $x \in A - B$ means $x \in A$ and $x \notin B$ - $x \in B - A$ means $x \in B$ and $x \notin A$ 8. **Therefore:** $$ (A \cup B) - (A \cap B) = (A - B) \cup (B - A) $$ **Final answer:** The equality is proven by set membership equivalence.