B-Algebra Power Law

Theorem
Let $\struct {X, \circ}$ 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 $\map P n$ 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
$\map P 1$ 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 $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ 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:
 * $\forall x \in X: \forall n, m \in \N: n \ge m \implies x^n \circ x^m = x^{n - m}$