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

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Theorem

Let $\left({X, \circ}\right)$ be a $B$-algebra.


Then:

$\forall x \in X: 0 \circ \left({0 \circ x}\right) = x$


Proof

Let $x \in X$.

Then:

\(\displaystyle 0 \circ x\) \(=\) \(\displaystyle \left({0 \circ x}\right)\circ 0\) Axiom $(A2)$ for $B$-algebras
\(\displaystyle \) \(=\) \(\displaystyle 0 \circ \left({0 \circ \left({0 \circ x}\right)}\right)\) Axiom $(A3)$ for $B$-algebras


From 0 in B-Algebra is Left Cancellable Element, we conclude:

$x = 0 \circ \left({0 \circ x}\right)$

as desired.

$\blacksquare$