Lemma 5

Suppose that the mapping $a$ is alternating and let $\mathbf{X}, \mathbf{Y}$ be two strongly associative subalgebras of $\mathbf{F}$. Then the subalgebra of $\mathbf{E}$ generated by $\mathbf{X} \cup \mathbf{Y}$ is associative.

Let $Z$ be the set of $z \in \mathbf{E}$ such that $a(u, v, z) = 0$ for all $u \in \mathbf{X}$ and $v \in \mathbf{Y}$, this is obviously a sub-A-module containing $\mathbf{X}$ and $\mathbf{Y}$ since $\mathbf{X}$ and $\mathbf{Y}$ are strongly associative; by Lemma 2, it will suffice to verify that, for $u \in \mathbf{X}$ and $v \in \mathbf{Y}$, $uZ \subset Z, vZ \subset Z, Zu \subset Z$ and $Zv \subset Z$. Now, for $u, u'$ in $\mathbf{X}$, $v \in \mathbf{Y}$ and $z \in Z$, by (2)

by virtue of the fact that $\mathbf{X}$ is strongly associative and the definition of $Z$. But as $\mathbf{X}$ is strongly associative, $a(u', u, zv) = 0$ and since $u'u \in \mathbf{X}$, $a(u'u, z, v) = 0$ by definition of $Z$. Hence $a(u', uz, v) = 0$, which shows that $uZ \subset Z$. Applying (2) now with $(p, q, r, s) = (v, z, u, u')$, we obtain similarly $Zu \subset Z$. Interchanging the roles of $\mathbf{X}$ and $\mathbf{Y}$ and using the fact that $a$ is alternating, we obtain $vZ \subset Z$ and $Zv \subset Z$; whence the lemma.

It now suffices, to prove that (b) implies (a) in Proposition 1, to take $\mathbf{X} = \{x\}$ and $\mathbf{Y} = \{y\}$, using Lemma 4.