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
Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n-m}$

Basis for the Induction
$P \left({1}\right)$ is true, as this just says:
 * $x \circ x = 0$

which follows from the definition of the zeroth power in $B$-algebra.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\forall m \in \N_{> 0}, m \le k: \forall x \in X: x^k \circ x^m = x^{k-m}$

Then we need to show:
 * $\forall m \in \N_{> 0}, m \le k+1: \forall x \in X: x^{k+1} \circ x^m = x^{k+1-m}$

Induction Step
This is our induction step:

First we show that:


 * $\forall x \in X: x^{k+1} \circ x = x^k$

Thus:

By induction, it follows that:
 * $\forall n \in \N_{>0}: \forall x \in X: x^n \circ x = x^{n-1}$

Now let $1 \le m \le k$.

We have:

By induction, it follows that:
 * $\forall x \in X: x^n \circ x^m = x^{n-m}$

for all $n, m \in N$ such that $n \ge m$.