Subset version of Artin's generalized theorem on alternative rings

Statement
Suppose $$R$$ is an alternative ring and $$X_1,X_2,X_3$$ are (possibly equal, possibly distinct) subsets of $$R$$ such that $$a(X_1,X_1,R) = a(X_2,X_2,R) = a(X_3,X_3,R) = a(X_1,X_2,X_3) = 0$$ where $$a$$ denotes the associator function (its application to a triple of sets simply means the set of possible outputs if the inputs are in the respective sets).

Then, the subring of $$R$$ generated by $$W = X_1 \cup X_2 \cup X_3$$ is an associative ring.

Applications

 * Artin's generalized theorem on alternative rings
 * Artin's theorem on alternative rings

Facts used

 * 1) In an alternative ring, the associator function is alternating in all pairs of variables.

Preliminary observations
To prove: $$a(W,W,W) = 0$$ where $$a$$ denotes the associator. Note that by $$0$$ we mean the singleton set $$\{ 0 \}$$.

Construction
As before, we denote the associator function by $$a$$. We use the letter $$f$$ to denote the Kleinfeld function. Define the following for subsets of $$R$$:

$$K := \{ x \in R \mid a(x,W,W) = f(x,W,W,W) = 0 \}$$

$$M := \{ m \in K \mid mK \subseteq K \mbox{ and } a(m,W,K) = 0 \}$$

$$S := \{ s \in M \mid a(s,M,K) = 0 \}$$

We will proceed to show that $$S$$ is an associative subring of $$R$$ that contains $$W$$. Note that this should suffice, even if $$S$$ is not the subring generated by $$W$$, because it would show that the subring generated by $$W$$, being contained inside $$S$$, must be associative.