B-Algebra Power Law

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

Let $n, m \in \N$ such that $n \ge m$.

Then:


 * $\forall x \in X: x^n \circ x^m = x^{n-m}$

where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.

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

Let $x \in X$.

Then:



Now showing: $\forall x \in X: n \in \N_{>0} \implies x^{n+1} \circ x = x^{n}$

Using a proof by induction: The base case of $n=1$ is established as true in the previous lemma. Assuming it is the case for some $k \in \N_{>0}$ we will show:


 * $x^{k+2} \circ x = x^{k+1}$:



Now proving the original proposition. Assuming for some $n, m \in \N_{>0}$ where $n-m \ge 1$:

First we establish: $x^n \circ x^{m+1} = x^{n-1} \circ x^m$

Using this result:

Hence the result.