0 in B-Algebra is Left Cancellable Element

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Theorem

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

Then:

$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$


Proof

Let $x, y \in X$ and let $0 \circ x = 0 \circ y$.

Then:

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


So we have shown:

$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x \circ y = 0$

From $B$-Algebra Identity: $x \circ y = 0 \iff x = y$:

$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$

Hence the result.

$\blacksquare$