1. **State the problem:** Prove that $ (A \cap C) \setminus ((A \setminus B) \setminus (B \setminus C)) = A \setminus B $ for all sets $A, B, C$. 2. **Recall set theory laws:** - Set difference: $X \setminus Y = X \cap Y^c$ - De Morgan's laws: $(X \cap Y)^c = X^c \cup Y^c$ - Distributive laws 3. **Rewrite the expression:** $$ (A \cap C) \setminus ((A \setminus B) \setminus (B \setminus C)) = (A \cap C) \cap \left[((A \setminus B) \setminus (B \setminus C))^c\right]$$ 4. **Express differences as intersections with complements:** $$A \setminus B = A \cap B^c,$$ $$B \setminus C = B \cap C^c,$$ thus $$ (A \setminus B) \setminus (B \setminus C) = (A \cap B^c) \setminus (B \cap C^c) = (A \cap B^c) \cap (B \cap C^c)^c.$$ 5. **Apply De Morgan's on the complement:** $$(B \cap C^c)^c = B^c \cup C,$$ so $$(A \setminus B) \setminus (B \setminus C) = (A \cap B^c) \cap (B^c \cup C) = (A \cap B^c \cap B^c) \cup (A \cap B^c \cap C).$$ 6. **Simplify:** $$A \cap B^c \cap B^c = A \cap B^c,$$ so we get $$(A \setminus B) \setminus (B \setminus C) = (A \cap B^c) \cup (A \cap B^c \cap C).$$ This is equivalent to $A \cap B^c$ since adding $A \cap B^c \cap C$ doesn't add new elements beyond $A \cap B^c$. 7. **Now substitute back:** $$(A \cap C) \setminus ((A \setminus B) \setminus (B \setminus C)) = (A \cap C) \cap (A \cap B^c)^c = (A \cap C) \cap (A^c \cup B).$$ 8. **Simplify:** $$(A \cap C) \cap (A^c \cup B) = [(A \cap C) \cap A^c] \cup [(A \cap C) \cap B] = \varnothing \cup (A \cap B \cap C) = A \cap B \cap C.$$ 9. **Comparing left and right hand sides:** Left is $A \cap B \cap C$ and right is $A \setminus B = A \cap B^c$. These are equal only if $B \cap C = \varnothing$, otherwise, the equality doesn't hold in general. 10. **Conclusion:** The expression $(A \cap C) \setminus ((A \setminus B) \setminus (B \setminus C)) = A \setminus B$ is not true for all sets $A, B, C$. For it to be true, additional conditions on $B$ and $C$ must hold. **Therefore, the given expression cannot be proven universally without additional constraints.**