# 0 in B-Algebra is Left Cancellable Element

## 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:

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

So we have shown:

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

Hence the result.

$\blacksquare$