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Let $S$ be a set such that if $A,B\in S$ then $A\cap B,A\triangle B\in S,$ where $\triangle$ denotes the symmetric difference operator. I would like to show that if $S$ contains $A$ and $B$, then it also contains $A\cup B, A\setminus B$.

The difference was easy to find, but I am not succeeding with the union. I was able to show that each of the sets $$\emptyset,A\setminus B, B\setminus A,(A\cap B)\cup(A\setminus B),(A\cap B)\cup(B\setminus A),A\cup(A\triangle B),B\cup(A\triangle B),\\(A\cap B)\cup(B\setminus A)\cup(A\triangle B)$$ is an element of $S$. Combining these I could go on producing other elements, and probably I would eventually find $A\cup B$. But I have already spent too long on the problem, there must be a cleverer approach, right?

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$$ \left(A\triangle B\right)\triangle\left(A\cap B\right)=A\cup B\;. $$

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    $\begingroup$ The most obvious... and the one I most seriously missed, right at the beginning. Silly me, thanks. $\endgroup$ – Richard May 27 '16 at 11:42
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$$A\cup B=((A \triangle B) \cap A)\triangle B$$

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Note that if $A\cap B=\varnothing$, then $A\cup B=A\mathbin\triangle B$. So if you can show that $A\setminus B\in S$, then $A\cup B=(A\setminus B)\mathbin\triangle B$.

Next, observe that $A\cap(A\mathbin\triangle B)=A\cap((A\setminus B)\cup(B\setminus A))=A\setminus B$.

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