Prove that $\mathcal{U}$ is a ring Let $X$ be a set. Then $\emptyset \neq \mathcal{S} \subset 2^{X}$ is called a semi-ring if:
(a) $A, B \in \mathcal{S} \Rightarrow A\cap B \in \mathcal{S}$
(b) $A, B \in \mathcal{S}$ imply that $A\setminus B$ is a finite disjoint union of elements of $\mathcal{S}$.
On the other hand, $\mathcal{S}$ is called a ring if $A, B \in \mathcal{S} \Rightarrow A \cup B \in \mathcal{S}$ and $A\setminus B \in \mathcal{S}$.
Let $\mathcal{S}$ be a semi-ring and $\mathcal{U}$ be the set of all finite disjoint unions of elements of $\mathcal{S}$. I want to show that $\mathcal{U}$ is a ring. It is straightfoward that $\mathcal{U}$ is closed under finite union so I only have to prove the second property.
My proof: Let $A, B \in \mathcal{U}$, so that:
$$A = \bigcup_{i=1}^{n}A_{i} \quad \mbox{and} \quad B = \bigcup_{j=1}^{m} B_{j}$$
where $A_{1},...,A_{n} \in \mathcal{S}$ are all disjoint and $B_{1},...,B_{m}\in \mathcal{S}$ are also disjoint. We have:
$$A \setminus B = \bigcup_{i=1}^{n}\bigcap_{j=1}^{m} A_{i}\setminus B_{j}$$
and because $\mathcal{S}$ is a semi-ring, every $A_{i}\setminus B_{j}$ is the finite union of elements of $\mathcal{S}$, say:
$$A_{i}\setminus B_{j} = \bigcup_{l=1}^{k(i,j)}F_{l}^{ij}$$
Hence,
$$A\setminus B = \bigcup_{i=1}^{n}\bigcap_{j=1}^{m}\bigcup_{l=1}^{k(i,j)}F_{l}^{ij} = \bigcup_{i=1}^{n}\bigcup_{l=1}^{k(i,j)}\bigcap_{j=1}^{m}F_{l}^{ij}$$
But since every $F_{l}^{ij} \in \mathcal{S}$ and $\mathcal{S}$ is closed under finite intersections, $S_{l}^{i} = \bigcap_{j=1}^{m}F_{l}^{ij} \in \mathcal{S}$ and these are all disjoint. Hence $A\setminus B \in \mathcal{U}$.
Is my proof correct? I'm not sure of these last arguments.
 A: I think your argument is correct. It took me a while to realize that you're rewriting $\bigcup_{l=1}^{k(i, j)}F_{l}^{ij}$ in order to directly enforce disjointness. I'm still not 100% sure what the function $k$ is.
I think you can show with a lemma that any finite not-necessarily-disjoint union in the semiring can be rewritten into a finite disjoint union. The union may grow in the number of components, but it will still be finite. If you prove that lemma separately, then you can directly use the original union of intersections.

I think you can also prove this with induction, specifically considering the number of components of the finite disjoint unions. I'm using some nonstandard notation here.
If $A_0$ is a non-empty finite disjoint union, then it is equal to $a:A$ where $a$ is an element of the union and $A$ is a possibly empty finite disjoint union. Using a $:$ this way is not standard notation. I'll also nonstandardly use $X^u$ to denote the union of the set $X$.
We can look at the different cases for $A^u \setminus B^u$.
$ \varnothing \setminus \varnothing $ is $\varnothing$.
$ A^u \setminus \varnothing $ is $A^u$, which is a finite disjoint union by hypothesis.
$ \varnothing \setminus B^u $ is $\varnothing$.
$ (a : A) \setminus (b : B) $ is $((a \setminus b) \cap (a \setminus B^u)) \cup ((A^u \setminus b) \cap (A^u \setminus B^u)) $. This is nearly the same as what you did, just packaged slightly differently. The outermost union is disjoint because $a$ and $A^u$ are disjoint by hypothesis and remain disjoint when elements are taken away.
We can show that our ring is closed under $(x, y) \mapsto x \setminus y$ by inducting on the length of the disjoint union with the most components.
