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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.

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    I think that to calculate $A \setminus (A \setminus B)$ is easier or at least looks nicer, but if you want to use logic, that's okay. Notice that $x \in A \setminus B$ reads "$x \in A$ but not $x \in B$" and $x \in A \cap B$ reads "$x \in A$ and $x \in B$". Clearly every point of $A$ either is or isn't in $B$, thus one of $A \setminus B$ and $A \cap B$ completely determines the other.2012-09-20

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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 $x \in A \wedge x \notin B$.

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    Yes. A\B is a$n$other way to say A-B.2012-09-20
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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.