0 in B-Algebra is Left Cancellable Element

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Theorem

Let $\left({X, \circ}\right)$ 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:

\(\displaystyle 0\) \(=\) \(\displaystyle x \circ x\) Axiom $(A1)$ for $B$-algebras
\(\displaystyle \) \(=\) \(\displaystyle \left({x \circ x}\right) \circ 0\) Axiom $(A2)$ for $B$-algebras
\(\displaystyle \) \(=\) \(\displaystyle x \circ \left({0 \circ \left({0 \circ x} \right)}\right)\) Axiom $(A3)$ for $B$-algebras
\(\displaystyle \) \(=\) \(\displaystyle x \circ \left({0 \circ \left({0 \circ y} \right)}\right)\) by hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left ({x \circ y}\right) \circ 0\) Axiom $(A3)$ for $B$-algebras
\(\displaystyle \) \(=\) \(\displaystyle x \circ y\) Axiom $(A2)$ 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.

$\blacksquare$