# B-Algebra Power Law with Zero

## Theorem

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

Let $n, m \in \N_{>0}$ such that $n > m$.

Then:

$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n-m}$

where $x^k$ for $k \in \N_{>0}$ denotes the $k$th power of the element $x$.

## Proof

First we show that: $\forall x \in X: x \circ x^2 = 0 \circ x$:

 $\ds x \circ x^2$ $=$ $\ds x \circ \left({x^1 \circ \left({0 \circ x}\right)}\right)$ Definition of B-Algebra Power of Element $\ds$ $=$ $\ds x \circ \left({x \circ \left({0 \circ x}\right)}\right)$ First Power of Element in B-Algebra $\ds$ $=$ $\ds \left({x \circ x}\right) \circ x$ Axiom $(A3)$ for $B$-algebras $\ds$ $=$ $\ds 0 \circ x$ Axiom $(A1)$ for $B$-algebras

$\Box$

Now we show that: $\forall x \in X: m \in \N_{>0}: x \circ x^m = 0 \circ x^{m-1}$:

 $\ds x \circ x^m$ $=$ $\ds x \circ \left({x^{m-1} \circ \left({0 \circ x}\right)}\right)$ Definition of B-Algebra Power of Element $\ds$ $=$ $\ds \left({x \circ x}\right) \circ x^{m-1}$ Axiom $(A3)$ for $B$-algebras $\ds$ $=$ $\ds 0 \circ x^{m-1}$ Axiom $(A1)$ for $B$-algebras

$\Box$

Now proving the original proposition using a proof by induction. The base case for $n=1, m=2$ was established in the first lemma.

Assuming $x^m \circ x^n = 0 \circ x^{n-m}$ for some $n, m \in N_{>0}$:

 $\ds x^{m+1} \circ x^n$ $=$ $\ds \left({x^m \circ \left({0 \circ x}\right)}\right) \circ x^n$ Definition of B-Algebra Power of Element $\ds$ $=$ $\ds x^m \circ \left({x^n \circ x}\right)$ Identity: $x \circ \left({y \circ z}\right) = \left({x \circ \left({0 \circ z}\right)}\right) \circ y$ $\ds$ $=$ $\ds x^m \circ x^{n-1}$ $B$-Algebra Power Law $\ds$ $=$ $\ds x^0 \circ x^{n-m-1}$ Induction using the previous lemma $\ds$ $=$ $\ds 0 \circ x^{n-m-1}$ Definition of Zeroth Power of Element

Hence the result.

$\blacksquare$