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.
The unknot defeats a universal coefficient-bound extension
Statement refuted
Assume full AC for the Jones invariant and for . A classical knot is a smooth embedding considered up to ambient isotopy. A regular diagram is a generic plane projection with only transverse double points, each marked with over/under information. The false universal extension is: for every classical knot , including the unknot, there is an integer with and . Here is the normalized library Jones polynomial with , is defined by the unique expansion , and is the least crossing number of a regular diagram of .
The boundary input is the unknot , represented by the crossing-free circle:
Statement references
Facts & Assumptions
Given: Full AC, , and the unknot with its one-component crossing-free diagram.
The library Jones polynomial has , and the coefficient vector is obtained from the unique finite expansion in powers of (Jones–Vassiliev coefficient vectors).
Under , is the least element of the set of crossing counts of regular diagrams representing (Crossing number of a link).
Full AC is the family-choice assertion (The Axiom of Choice).
selects an element from each countable family of nonempty sets (The Axiom of Countable Choice ()).
Counterexample
Under [A1], by [F1]. Its unique coefficient expansion is , , and for every . Thus for every positive order.
The displayed diagram has zero crossings, so under [A2] the set in [F2] contains ; all crossing counts are nonnegative. Hence , and there is no integer satisfying .
The proposed universal statement includes , but steps 1.1 and 1.2 show that no required nonzero vector or admissible order exists for it. Therefore the statement is false.
Remarks
The source's actual coefficient claim assumes that is nontrivial, so this counterexample refutes only the broader universal extension. The presence or absence of DOI or journal metadata is bibliographic information; by itself it proves neither the source theorem nor its negation.
Remarks references
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
19 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
- Avishy Carmi and Eliahu Cohen, On trivial Jones–Vassiliev polynomials, arXiv:2606.22410v1, submitted 2026-06-21 (standard reference, not scraped)
- Justin Roberts, Knots Knotes, §4.4.8 (standard reference, not scraped)