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 $x$.

$B$-Algebra Power Law with Zero

$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$

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:

\(\ds x^{k + 1} \circ x\)

\(=\)

\(\ds \paren {x^k \circ \paren {0 \circ x} } \circ x\)

Definition of $B$-Algebra Power of Element

\(\ds \)

\(=\)

\(\ds x^k \circ \paren {x \circ \paren {0 \circ \paren {0 \circ x} } }\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x^k \circ \paren {\paren {x \circ x} \circ 0}\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x^k \circ \paren {x \circ x}\)

$B$-Algebra Axiom $(\text A 2)$

\(\ds \)

\(=\)

\(\ds x^k \circ 0\)

$B$-Algebra Axiom $(\text A 1)$

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds x^{k + 1} \circ x\)

\(=\)

\(\ds x^k\)

$B$-Algebra Axiom $(\text A 2)$

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:

\(\ds x^{k + 1} \circ x^{m + 1}\)

\(=\)

\(\ds x^{k + 1} \circ \paren {x^m \circ \paren {0 \circ x} }\)

Definition of $B$-Algebra Power of Element

\(\ds \)

\(=\)

\(\ds \paren {x^{k + 1} \circ x} \circ x^m\)

$B$-Algebra Axiom $(\text A 3)$

\(\ds \)

\(=\)

\(\ds x^k \circ x^m\)

from $(1)$

\(\ds \)

\(=\)

\(\ds x^{k - m}\)

Induction Hypothesis

By induction, it follows that:

$\forall x \in X: \forall n, m \in \N: n \ge m \implies x^n \circ x^m = x^{n - m}$

$\blacksquare$

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