B-Algebra is Left Cancellable

Theorem

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

Then $\circ$ is a left cancellable operation.

Proof

Let $x, y, z \in X$:

\(\ds x \circ y\)

\(=\)

\(\ds x \circ z\)

\(\ds \leadstoandfrom \ \ \)

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

\(=\)

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

$0$ in $B$-Algebra is Left Cancellable Element

\(\ds \leadstoandfrom \ \ \)

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

\(=\)

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

Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$

\(\ds \leadstoandfrom \ \ \)

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

\(=\)

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

$0$ in $B$-Algebra is Left Cancellable Element

\(\ds \leadstoandfrom \ \ \)

\(\ds y \circ x\)

\(=\)

\(\ds z \circ x\)

$0$ in $B$-Algebra is Left Cancellable Element

\(\ds \leadstoandfrom \ \ \)

\(\ds y\)

\(=\)

\(\ds z\)

$B$-Algebra is Right Cancellable

Hence the result.

$\blacksquare$