ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Dihedral actions of prime and composite degree

Example

Let $D_{n}$ act on the vertices of a regular $n$ -gon, identified with $Z / n Z$ .

If $n$ is prime, this action is primitive. If $n$ is composite, then for every divisor $d$ with $1 < d < n$ the congruence classes modulo $d$ form a nontrivial block system.

Facts & Assumptions

Given: The natural action of the dihedral group $D_{n}$ on the vertices $Z / n Z$ of the regular $n$ -gon.

[L1]

A transitive action of prime degree is primitive (A transitive action of prime degree is primitive).

[L2]

A block is a nonempty subset $B$ such that for every group element $g$ , either $g \cdot B = B$ or $(g \cdot B) \cap B = \emptyset$ (Blocks and block systems for a group action).

Verification

technique · direct

1.1

The action of $D_{n}$ on the $n$ vertices is transitive, so if $n$ is prime, [L1] makes it primitive.

1.2

Suppose $n$ is composite and let $d$ satisfy $1 < d < n$ and $d ∣ n$ . Put $B : = {0, d, 2 d, \dots, n - d} \subseteq Z / n Z$ . Rotations send $B$ to its residue-class translates modulo $d$ , and reflections send residue classes modulo $d$ to residue classes modulo $d$ as well. Hence every dihedral image of $B$ is either $B$ itself or a disjoint residue class, so [L2] makes $B$ a block.

2.1

Because $1 < d < n$ , the block $B$ is neither a singleton nor all of $Z / n Z$ . So composite degree produces nontrivial blocks.

step 1.2∎

Depends on

Used by

FALSE: every block is an invariant subset False statement

Dependency tree · two levels

6 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

J. S. Milne, Group Theory, Chapter 4 (standard reference, not scraped)
K. Conrad, Transitive Group Actions (standard reference, not scraped)