1. **State the problem:** Prove that $ (A \cup B) - (A \cap B) = (A - B) \cup (B - A) $. 2. **Recall definitions:** - $A \cup B$ is the set of elements in $A$ or $B$ or both. - $A \cap B$ is the set of elements in both $A$ and $B$. - $A - B$ is the set of elements in $A$ but not in $B$. 3. **Rewrite the left side:** $$ (A \cup B) - (A \cap B) = \{x \mid x \in (A \cup B) \text{ and } x \notin (A \cap B)\} $$ This means $x$ is in $A$ or $B$, but not in both. 4. **Rewrite the right side:** $$ (A - B) \cup (B - A) = \{x \mid (x \in A \text{ and } x \notin B) \text{ or } (x \in B \text{ and } x \notin A)\} $$ This means $x$ is in exactly one of the sets $A$ or $B$. 5. **Show both sides describe the same set:** - If $x \in (A \cup B) - (A \cap B)$, then $x$ is in $A$ or $B$ but not both, so $x$ is in $A$ and not $B$, or in $B$ and not $A$. - Conversely, if $x \in (A - B) \cup (B - A)$, then $x$ is in $A$ or $B$ but not both, so $x$ is in $A \cup B$ but not in $A \cap B$. 6. **Conclusion:** Since both sides contain exactly the elements that are in one of the sets but not both, we have $$ (A \cup B) - (A \cap B) = (A - B) \cup (B - A) $$ which completes the proof.