B-Algebra is Right Cancellable
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Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is right-cancellable for $X$. That is:
- $\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$
Proof
Let $x, y \in X$.
Then:
\(\ds \paren {x \circ y} \circ \paren {0 \circ y}\) | \(=\) | \(\ds x \circ \paren {\paren {0 \circ y} \circ \paren {0 \circ y} }\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ 0\) | $B$-Algebra Axiom $(\text A 1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) | $B$-Algebra Axiom $(\text A 2)$ |
Now if for some $z \in X$ we have $x \circ z = y \circ z$, then also:
- $\paren {x \circ z} \circ \paren {0 \circ z} = \paren {y \circ z} \circ \paren {0 \circ z}$
which by the above implies $x = y$.
Hence the result.
$\blacksquare$