Commutative B-Algebra Implies (zy)(zx)=xy

Theorem

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

Then:

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

Proof

Let $x, y, z \in X$.

Then:

\(\ds \paren {z \circ y} \circ \paren {z \circ x}\)

\(=\)

\(\ds \paren {\paren {z \circ y} \circ \paren {0 \circ x} } \circ z\)

Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$

\(\ds \)

\(=\)

\(\ds \paren {x \circ \paren {0 \circ \paren {z \circ y} } } \circ z\)

Definition of Commutative $B$-Algebra

\(\ds \)

\(=\)

\(\ds x \circ \paren {z \circ \paren {0 \circ \paren {0 \circ \paren {z \circ y} } } }\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x \circ \paren {z \circ \paren {z \circ y} }\)

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

\(\ds \)

\(=\)

\(\ds x \circ y\)

$B$-Algebra is Commutative iff $x \circ \paren {x \circ y} = y$

$\blacksquare$