Product of Powers in B-Algebra

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

Let $x \in X$ and $m, n \in \N$.

Then:


 * $x^m \circ x^n = \begin{cases}

x^{m-n} & : m \ge n \\ 0 \circ x^{n-m} & : n > m \end{cases}$

Proof
For $m \ge n$ the result follows from $B$-Algebra Power Law.

For $n > m$ the result follows from $B$-Algebra Power Law with Zero.