B-Algebra Power Law with Zero

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Theorem

Let $\struct {X, \circ}$ 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:

$\forall x \in X: x \circ x^2 = 0 \circ x$
\(\ds x \circ x^2\) \(=\) \(\ds x \circ \paren {x^1 \circ \paren {0 \circ x} }\) Definition of Power ($B$-Algebra)
\(\ds \) \(=\) \(\ds x \circ \paren {x \circ \paren {0 \circ x} }\) First Power of Element in $B$-Algebra
\(\ds \) \(=\) \(\ds \paren {x \circ x} \circ x\) $B$-Algebra Axiom $(\text A 3)$
\(\ds \) \(=\) \(\ds 0 \circ x\) $B$-Algebra Axiom $(\text A 1)$

$\Box$


Now we show that:

$\forall x \in X: m \in \N_{>0}: x \circ x^m = 0 \circ x^{m - 1}$
\(\ds x \circ x^m\) \(=\) \(\ds x \circ \paren {x^{m - 1} \circ \paren {0 \circ x} }\) Definition of Power ($B$-Algebra)
\(\ds \) \(=\) \(\ds \paren {x \circ x} \circ x^{m - 1}\) $B$-Algebra Axiom $(\text A 3)$
\(\ds \) \(=\) \(\ds 0 \circ x^{m - 1}\) $B$-Algebra Axiom $(\text A 1)$

$\Box$


Now we prove the original proposition using a proof by induction.

The base case for $n = 1, m = 2$ was established in the first lemma.


Let us assume $x^m \circ x^n = 0 \circ x^{n - m}$ for some $n, m \in \N_{>0}$.

Then:

\(\ds x^{m + 1} \circ x^n\) \(=\) \(\ds \paren {x^m \circ \paren {0 \circ x} } \circ x^n\) Definition of Power ($B$-Algebra)
\(\ds \) \(=\) \(\ds x^m \circ \paren {x^n \circ x}\) Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
\(\ds \) \(=\) \(\ds x^m \circ x^{n - 1}\) $B$-Algebra Power Law
\(\ds \) \(=\) \(\ds x^0 \circ x^{n - m - 1}\) Induction using the previous lemma
\(\ds \) \(=\) \(\ds 0 \circ x^{n - m - 1}\) Definition of Zeroth Power of Element

Hence the result.

$\blacksquare$