Sum of Ring Products is Subring of Commutative Ring

Theorem
Let $$\left({R, +, \circ}\right)$$ be a commutative ring.

Let $$\left({S, +, \circ}\right)$$ and $$\left({T, +, \circ}\right)$$ be subrings of $$\left({R, +, \circ}\right)$$.

Let $$S T$$ be defined as:
 * $$S T = \left\{{\sum_{i=1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1\,.\,.\ n}\right]}\right\}$$

Then $$S T$$ is a subring of $$\left({R, +, \circ}\right)$$.

Proof
From Sum of All Ring Products is Additive Subgroup we have that $$\left({S T, +}\right)$$ is an additive subgroup of $$R$$.

Let $$x_1, x_2 \in S T$$.

Then:
 * $$x_1 = \sum_{i=1}^m s_i \circ t_i, x_2 = \sum_{i=1}^n s_j \circ t_j$$

for some $$s_i, t_i, s_j, t_j, m, n$$, etc.

Then:

$$ $$

So $$x_1 \circ x_2$$ and the result follows from the Subring Test.