Sum of All Ring Products is Associative

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

Let $$\left({S, +}\right), \left({T, +}\right), \left({U, +}\right)$$ be additive subgroups 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:
 * $$\left({S T}\right) U = S \left({T U}\right)$$

Proof
We have by definition that $$S T$$ is made up of all finite sums of elements of the form $$s \circ t$$ where $$s \in S, t \in T$$.

From Sum of All Ring Products is Closed under Addition, this set is closed under ring addition.

Therefore, so are $$\left({S T}\right) U$$ and $$S \left({T U}\right)$$.

Let $$z \in \left({S T}\right) U$$.

Then $$z$$ is a finite sum of elements in the form $$x \circ u$$ where $$x \in ST$$ and $$u \in U$$.

So $$x$$ is a finite sum of elements in the form $$s \circ t$$ where $$s \in S, t \in T$$.

Therefore $$z$$ is a finite sum of elements in the form $$\left({s \circ t}\right) \circ u$$ where $$s \in S, t \in T, u \in U$$.

As $$\left({R, +, \circ}\right)$$ os a ring, $$\circ$$ is associative.

So $$z$$ is a finite sum of elements in the form $$s \circ \left({t \circ u}\right)$$ where $$s \in S, t \in T, u \in U$$.

So these elements all belong to $$S \left({T U}\right)$$.

Since $$S \left({T U}\right)$$ is closed under addition, $$z \in S \left({T U}\right)$$.

So:
 * $$\left({S T}\right) U \subseteq S \left({T U}\right)$$

By a similar argument in the other direction, $$S \left({T U}\right) \subseteq \left({S T}\right) U $$

and so:
 * $$\left({S T}\right) U = S \left({T U}\right)$$

by Equality of Sets.