Flexible algebra

Remarks

In terms of the associator, the condition is $(x, y, x) = 0$ for all $x, y$. Example: every commutative algebra is flexible. Indeed $(xy)x = x(xy)$ by commutativity applied to $xy$ and $x$, and $x(yx) = x(xy)$ by commutativity applied to $y$ and $x$. Non-example: let $A$ have basis $e, f$ over a field, with $ee = f$, $ef = e$ and $fe = ff = 0$. Then $(ee)e = fe = 0$, while $e(ee) = ef = e$, so the condition fails for $x = y = e$.