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: $\forall x \in X: x \circ x^2 = 0 \circ x$:



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



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 $n, m \in N_{>0}$:

Hence the result.