Prove or disprove that if $A\setminus B = A\setminus C$ then $A \cap B = A \cap C$ I've been having trouble solving this problem and I have no clue where to go at this point. If anyone could help me out and explain along the way I'd appreciate it greatly. 
Let $A,B,C$ be a sets.  Supppose that $A\setminus B = A\setminus C$, then $A \cap B = A \cap C$. Prove or disprove with a counterexample. 
 A: Hint. Note that $(A \setminus B) \cup (A \cap B) = A$. Note, also, that the two sets are disjoint. Try writing things in terms of logic, if you don't see intuitively. $x \in A \setminus B$ means $x \in A \wedge x \notin B$.
A: Expanding $A \setminus B = A \setminus C$ using the definition $$x \in A \setminus B \;\equiv\; x \in A \land \lnot(x \in B)$$ and then simplifying results in
$$
\begin{align}
& A \setminus B = A \setminus C \\
\equiv & \;\;\;\;\;\text{"extensionality"} \\
& \langle \forall x :: x \in A \setminus B \equiv x \in A \setminus C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\setminus$, twice"} \\
& \langle \forall x :: x \in A \land \lnot(x \in B) \equiv x \in A \land \lnot(x \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: move common conjunct out of $\equiv$"} \\
& \langle \forall x :: x \in A \Rightarrow (\lnot(x \in B) \equiv \lnot(x \in C)) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify by negating both sides of $\equiv$"} \\
& \langle \forall x :: x \in A \Rightarrow (x \in B \equiv x \in C) \rangle \\
\end{align}
$$
Now, in the same way expand $A \cap B = A \cap C$ using the definition $$x \in A \cap B \;\equiv\; x \in A \land x \in B$$ and compare the results.
