B-Algebra Identity: x (y z) = (x (0 z)) y

Theorem

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

Then:

$\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$

Proof

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

Then:

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

\(=\)

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

$B$-Algebra Axiom $(\text A 2)$

\(\ds \)

\(=\)

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

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

Hence the result.

$\blacksquare$

Also see

This identity is comparable to $B$-Algebra Axiom $(\text A 3)$.