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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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    Have you tried drawing a picture?2012-09-20
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    I know that it takes into consideration everything that's in A but not in B and everything that's in A but not in C. With this taken into consideration, B and C definitely don't have to be the same set and A\B can still be equal to A\C. Therefore, if A\B=A\C, then A∩B=A∩C. I'm pretty sure that this is right but I'm not sure how to actually make a proof out of it.2012-09-20
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    Keep in mind that I just started writing proofs a week ago.2012-09-20
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    Although an "if ... then ..." proof might be okay, I suggest you evaluate $A \setminus (A \setminus B)$. I assume you know how to write difference as intersection and a complement of intersection as union.2012-09-20
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    So I should start the proof off by restating the problem and then also saying "This means that the is an element x \in A that is not in B and an element x \in A that is also not in C. Suppose x \in B and x \in C." and then continue to prove by contradiction? Would this be the best way to go about it?2012-09-20
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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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    What does the arrow cone shaped ^ mean?2012-09-20
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    $\wedge$ stands for **and**. $\vee$ stands for **or** and $\neg$ stands for **not**. It is useful to know that set operations correspond to logical ones. $\cup$ is $\vee$, $\cap$ is $\wedge$ and complement is $\neg$.2012-09-20
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    I haven't been taught that unfortunately but I have been taught the not symbols. Thank you for the information!2012-09-20
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    @KarolisJuodelė Does A/B another way of saying A-B?2012-09-20
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    Yes. A\B is another 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.