Ring of Sets Generated by Semiring

Theorem
Let $$\mathcal S$$ be a semiring of sets.

Let $$\mathcal R \left({\mathcal S}\right)$$ be the minimal ring generated by $\mathcal S$.

Let $$\mathcal L$$ be the system of sets $$A$$ with the finite expansions:
 * $$A = \bigcup_{k=1}^n A_k$$

with respect to the sets $$A_k \in \mathcal S$$.

Then $$\mathcal L = \mathcal R \left({\mathcal S}\right)$$.

Proof
First we need to show that $$\mathcal L$$ is a ring of sets.

Let $$A, B \in \mathcal L$$.

Then by definition of $$\mathcal L$$, they have expansions:

$$ $$

Since $$\mathcal S$$ is a semiring of sets, we have:
 * $$C_{ij} = A_i \cap B_j \in \mathcal S$$

By Pairwise Disjoint Subsets in Semiring Part of Partition, there exist finite expansions:

$$ $$

From these, it follows that $$A \cap B$$ and $$A \ast B$$ have the finite expansions:

$$ $$

Hence both $$A \cap B \in \mathcal L$$ and $$A \ast B \in \mathcal L$$.

So by definition, $$\mathcal L$$ is a ring of sets.

From the details of the above construction, the fact that $$\mathcal L$$ is the minimal ring generated by $$\mathcal S$$ follows immediately.