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

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Theorem

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


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

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


Proof

Necessary Condition

Let $x, y \in X$:

\(\ds x \circ \paren {x \circ y}\) \(=\) \(\ds \paren {x \circ \paren {0 \circ y} } \circ x\) Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
\(\ds \) \(=\) \(\ds \paren {y \circ \paren {0 \circ x} } \circ x\) $\struct {X, \circ}$ is commutative
\(\ds \) \(=\) \(\ds y \circ \paren {x \circ x}\) $B$-Algebra Axiom $(\text A 3)$
\(\ds \) \(=\) \(\ds y \circ 0\) $B$-Algebra Axiom $(\text A 1)$
\(\ds \) \(=\) \(\ds y\) $B$-Algebra Axiom $(\text A 2)$

$\Box$


Sufficient Condition

Let $x, y \in X$:

\(\ds \paren {x \circ \paren {0 \circ y} } \circ y\) \(=\) \(\ds x \circ \paren {y \circ \paren {0 \circ \paren {0 \circ y} } }\) $B$-Algebra Axiom $(\text A 3)$
\(\ds \) \(=\) \(\ds x \circ \paren {y \circ y}\) Identity: $x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$
\(\ds \) \(=\) \(\ds y \circ \paren {x \circ x}\) $B$-Algebra Axiom $(\text A 3)$
\(\ds \) \(=\) \(\ds x \circ 0\) $B$-Algebra Axiom $(\text A 2)$
\(\ds \) \(=\) \(\ds x\) $B$-Algebra Axiom $(\text A 1)$
\(\ds \) \(=\) \(\ds y \circ \paren {y \circ x}\) by hypothesis
\(\ds \) \(=\) \(\ds y \circ \paren {y \circ \paren {0 \circ \paren {0 \circ x} } }\) Identity: $x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$
\(\ds \) \(=\) \(\ds \paren {y \circ \paren {0 \circ x} } \circ y\) $B$-Algebra Axiom $(\text A 3)$


Hence:

$\paren {x \circ \paren {0 \circ y} } \circ y = \paren {y \circ \paren {0 \circ x} } \circ y$

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

$x \circ \paren {0 \circ y} = y \circ \paren {0 \circ x}$

and hence commutativity.

$\blacksquare$

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