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

Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.

Then $\struct {X, \circ}$ is commutative :


 * $\forall x, y \in X: x \circ \paren {x \circ y} = y$

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 commutativity.