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

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Theorem

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


Then:

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


Proof

Let $x, y \in X$.

Then:

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


Hence the result.

$\blacksquare$