Lemma 3

Suppose that the mapping $a$ is alternating. If $H$ is a strongly associative subset of $F$, the subalgebra of $F$ generated by $H$ is strongly associative.

As the set of strongly associative subsets of $F$ is inductive, it suffices to prove that if $H$ is a maximal strongly associative subset of $F$, $H$ is then a subalgebra of $F$. As $H$ is obviously a sub-A-module of $F$, it suffices to verify that for any two elements $u, v$ of $H$, $H \cup \{uv\}$ is also strongly associative, for by virtue of the definition of $H$, this will imply $uv \in H$. Now, for all $z \in H$ and all $t \in F$, by (2)

since $\mathbf{H}$ is strongly associative; as $u, v, z$ are in $\mathbf{H}$, also

whence $a(uv, t, z) = 0$. Using the fact that $a$ is alternating, this shows that $a(p, q, r) = 0$ whenever at least two of the elements $p, q, r$ belong to $\mathbf{H} \cup \{uv\}$ whence the lemma.