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 $$A_2 = A - A_1$$, where $$A - A_1$$ 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 $$A - A_1 = A_2 \in \mathcal {R}$$.

But by Set Difference with Intersection, we have that $$\left({A - A_1}\right) \cap A_1 = \varnothing$$.

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.