# Set theory: what is other way to represent A\B?

I try to represent basic set operations using other operations with some limitations. For example $A \setminus B$ using only $\cup$ and $\oplus$ (Symmetric difference):

$A \setminus B = A \oplus (A \oplus B) \oplus (A \cup B)$

But I stuck with the same set operation but using onlu $\oplus$ and $\cap$.

I came to something like: $A \setminus B = (A \oplus B) \oplus (A \cap B) \cap B$

Is it correct or did I miss something?

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What operation is $\oplus$? – Enjoys Math Oct 28 '13 at 21:45
@Dragon It may be confusing to talk about XOR when working with sets. Perhaps you may want to say "symmetric difference" if that fits your operation. en.wikipedia.org/wiki/Symmetric_difference – Lord Soth Oct 28 '13 at 21:54
@LordSoth, yes, you're right. Sorry for confusing. – Dragon Oct 28 '13 at 21:58
Your expression for $A\setminus B$ is patently more complicated that it needs to be, because $A\oplus(A\oplus B)=(A\oplus A)\oplus B=\emptyset\oplus B=B$ – bof Oct 28 '13 at 22:00
@bof, I knew I did something wrong. I even didn't manage to describe my expression with Venn diagram. Now it seems I understood my fault. – Dragon Oct 28 '13 at 22:05

$A\setminus B=B\oplus(A\cup B)=A\oplus(A\cap B)$

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Thanks, it looks simple enough. Maybe I had to draw Venn diagram first. – Dragon Oct 29 '13 at 9:02

Let $x \in A$ be a predicate and $x \in B$ be a predicate and form the table

$$\begin{vmatrix} x\in A & x \in B & x \in A - B \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{vmatrix}$$

Let the variable $y$ take the place of $x \in A$, and $z$ the place of $x \in B$. Then the predicate $p = x \in A - B$ is the polynomial $p(y, z) = y \bar{z}$, constructed by examining the table. This translates back to sets as $A - B = A \cap \bar{B}$.

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But this one doesn't use symmetric difference – Dragon Oct 29 '13 at 9:01