Artin's Theorem on Alternative Algebras

Theorem
Let $A = \left({A_R, \oplus}\right)$ be an algebra over the ring $R$ such that $A$ is not a boolean algebra.

Then $A$ is alternative iff:
 * $\forall a, b \in A: \left({a \oplus a}\right) \oplus b = a \oplus \left({a \oplus b}\right)$
 * $\forall a, b \in A: \left({b \oplus a}\right) \oplus a = b \oplus \left({a \oplus a}\right)$

Alternative Definition
Some sources take this as a definition of an alternative algebra and from it deduce that $A$ is alternative iff:
 * for all $a, b \in A_R$, the subalgebra generated by $\left\{{a, b}\right\}$ is an associative algebra.

However, the latter statement is how an alternative algebra is defined on this site, and from it we deduce the statements given in the exposition of this theorem.

The approaches are clearly equivalent.

Proof
Suppose that:
 * $\forall a, b \in A: \left({a \oplus a}\right) \oplus b = a \oplus \left({a \oplus b}\right)$
 * $\forall a, b \in A: \left({b \oplus a}\right) \oplus a = b \oplus \left({a \oplus a}\right)$

Then:
 * $\left[{a, a, b}\right] = 0$
 * $\left[{b, a, a}\right] = 0$

where $\left[{a, a, b}\right]$ denotes the associator of $a, b \in A_R$.