1. **State the problem:** Prove the set equality $ (A - B) - (B - C) = A - B $ for all sets $ A $, and given sets $ B $ and $ C $. 2. **Recall definitions and laws:** - Set difference: $ X - Y = \{x \mid x \in X \text{ and } x \notin Y\} $. - We will use laws like distributivity, association, and subset arguments. 3. **Rewrite the left-hand side (LHS):** $$ (A - B) - (B - C) = \{x \mid x \in A - B \text{ and } x \notin B - C\} $$ 4. **Expand the definitions inside:** - $x \in A - B$ means $x \in A$ and $x \notin B$. - $x \notin B - C$ means $x \notin \{y \mid y \in B \text{ and } y \notin C\}$, i.e., $x \notin B$ or $x \in C$ (by De Morgan's law applied to $B - C$). 5. **Combine conditions for $x$ in LHS:** $$ x \in A, x \notin B, \text{ and } (x \notin B \text{ or } x \in C) $$ Since $x \notin B$ is already true, the disjunction $x \notin B \text{ or } x \in C$ is automatically true. 6. **Simplify:** The whole condition reduces to: $$ x \in A \text{ and } x \notin B $$ which by definition is $x \in A - B$. 7. **Therefore:** $$ (A - B) - (B - C) = A - B $$ 8. **Conclusion:** The identity holds for all sets $A$, using the definitions of set difference and basic logic laws like De Morgan's and distributivity.