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?