Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29
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Every null word has a minimal algebraic relator area

Statement

Let P be a finite presentation and let w be trivial in the group presented by P. Then AreaP(w) exists.

Facts & Assumptions

Given: A finite presentation P and a word w with w=P1.

[L1]

The algebraic relator area of a null word is defined as the least length of a relator expression for that word. (Algebraic relator area and the Dehn function of a finite presentation)

Proof

technique · direct
1.1

Because w is null, the admissible lengths in [L1] form a nonempty subset of N: every relator expression for w contributes one such length, and the empty product contributes the value 0 in the boundary case.

L1given
2.1

Every nonempty subset of N has a least element. Applying this to the set of admissible lengths from step 1.1 gives a least m, and [L1] defines that least number to be AreaP(w).

L1step 1.1
3.1

Hence the minimal algebraic relator area exists for every null word.

step 2.1

Depends on

Used by

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources