B-Algebra is Commutative iff x(xy)=y

Theorem
Let $\left({X, \circ}\right)$ be a $B$-algebra.

Then $\left({X, \circ}\right)$ is $0$-commutative iff:


 * $\forall x, y \in X: x \circ \left({x \circ y}\right) = y$

Necessary Condition
Let $x, y \in X$:

Sufficient Condition
Let $x, y \in X$:

Hence:


 * $\left({x \circ \left({0 \circ y} \right)} \right) \circ y = \left ({y \circ \left({0 \circ x} \right)} \right) \circ y$

From $B$-Algebra is Right Cancellable, we have:


 * $x \circ \left({0 \circ y} \right) = y \circ \left({0 \circ x} \right)$

and hence $0$-commutativity.