B-Algebra Identity: 0(0x)=x

Theorem

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

Then:

$\forall x \in X: 0 \circ \paren {0 \circ x} = x$

Proof

Let $x \in X$.

Then:

\(\ds 0 \circ x\)

\(=\)

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

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

\(\ds \)

\(=\)

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

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

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

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

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

$\blacksquare$