Ring of Sets is Semiring of Sets

Theorem
Let $$\mathcal R$$ be a ring of sets.

Then $$\mathcal R$$ is also a semiring of sets.

Proof
Let $$A \in \mathcal R$$.

Suppose $$A_1 \subseteq A$$.

Let $$A_2 = A \setminus A_1$$, where $$A \setminus A_1$$ denotes set difference.

By definition, $$A_2$$ is then the relative complement of $$A_1$$ with respect to $$A$$.

From Union with Relative Complement it then follows that $$A_1 \cup A_2 = A$$.

By Ring of Sets Closed under Various Operations, we have that $$A \setminus A_1 = A_2 \in \mathcal R$$.

But by Set Difference Intersection Second Set is Null, we have that $$\left({A \setminus A_1}\right) \cap A_1 = \varnothing$$.

Hence for any given $$A$$ for which we have $$A_1 \subseteq A$$, we can represent $$A$$ as the finite expansion $$A_1 \cup A_2$$ such that $$A_1 \cap A_2 = \varnothing$$ and $$A_1, A_2 \in \mathcal R$$

Thus, by definition, $$\mathcal R$$ is a semiring of sets.