0 in B-Algebra is Left Cancellable Element

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:

\(\ds 0\)

\(=\)

\(\ds x \circ x\)

$B$-Algebra Axiom $(\text A 1)$

\(\ds \)

\(=\)

\(\ds \paren {x \circ x} \circ 0\)

$B$-Algebra Axiom $(\text A 2)$

\(\ds \)

\(=\)

\(\ds x \circ \paren {0 \circ \paren {0 \circ x} }\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x \circ \paren {0 \circ \paren {0 \circ y} }\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \paren {x \circ y} \circ 0\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x \circ y\)

$B$-Algebra Axiom $(\text A 2)$

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.

$\blacksquare$