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.
$\blacksquare$