Sum of All Ring Products is Additive Subgroup

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

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

Let $$S + T$$ be defined as subset product.

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 both $$S + T$$ and $$S T$$ are additive subgroups of $$\left({R, +, \circ}\right)$$.

Proof
As $$\left({R, +}\right)$$ is abelian (from the definition of a ring), we have:
 * $$S + T = T + S$$

from Subset Product of Commutative is Commutative.

So from Subset Product of Subgroups it follows that $$S + T$$ is an additive subgroup of $$\left({R, +, \circ}\right)$$.

Let $$x, y \in S T$$.

We have that $\left({S T, +}\right)$ is closed.

So $$x + y \in S T$$.

So, if $$y = \sum s_i \circ t_i \in S T$$, it follows that $$-y = \sum \left({-s_i}\right) \circ t_i \in S T$$.

By the Two-Step Subgroup Test, we have that $$S T$$ is an additive subgroup of $$\left({R, +, \circ}\right)$$.