ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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 degree-one derivative of linking number

Example

Assume ${A C}_{ω}$ . For an oriented two-component link $L = L_{1} \cup L_{2}$ , let $lk (L_{1}, L_{2})$ be one half the sum of the signs of all crossings between distinct components, using $sgn \det (v_{o v e r}, v_{u n d e r})$ in the oriented projection plane with positive viewing normal. Extend the singular finite-difference formula to links by resolving marked crossings independently. At one marked mixed crossing, the positive-minus-negative derivative of $lk$ is $+ 1$ ; reversing the order of those two resolutions gives $- 1$ . A marked self-crossing has derivative $0$ . At any two distinct marked crossings, mixed or self, the second derivative is $0$ , so linking number is a type-at-most-one invariant of two-component links and, by the nonzero first derivative, has exact type one. The total pairwise linking number of a one-component knot is $0$ , because the defining sum over distinct component pairs is empty.

Example references

Bar-Natan and Stoimenow, §1.1, pp. 3–5

Chmutov, Duzhin, and Mostovoy, §§2.2.1 and 3.2.2, pp. 33–34 and 86

Facts & Assumptions

Given: The crossing-sign convention and the two-component oriented-link linking-number formula in the Example statement. The topological invariance of this quantity uses the separate ${A C}_{ω}$ supplier branch.

[F1]

Singular knots and Vassiliev derivatives: link finite differences are the alternating sums over independent marked-crossing resolutions.

[F2]

Oriented crossing signs, diagram writhe, and blackboard framing: a crossing sign is $sgn \det (v_{o v e r}, v_{u n d e r})$ ; a positive crossing has sign $+ 1$ and a negative crossing has sign $- 1$ .

[F3]

Compute oriented linking number from a diagram: on a two-component diagram the linking-number algorithm returns one half of the signed sum of mixed crossings.

[F4]

The diagram linking-number scan terminates and returns the stated integer: under ${A C}_{ω}$ , the finite diagram scan terminates and returns the integer linking number.

[F5]

Linking number is invariant and writhe changes under a kink: under ${A C}_{ω}$ , the algorithmic linking number is unchanged by oriented Reidemeister moves and hence is invariant under ambient isotopy.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ asserts that every sequence of nonempty sets has a choice function; only the topological supplier chain [F4]–[F5] uses it.

Verification

1.1givenA1F1F2F3F4F5algebra

Let $C$ be half the signed sum of all mixed crossings except the marked one. Under [A1], [F4] identifies the diagram half-sum with the integer linking number and [F5] makes it an ambient-isotopy invariant. By [F2] and [F3], the positive and negative resolutions contribute respectively $+ \frac{1}{2}$ and $- \frac{1}{2}$ to this linking number, so $lk (L_{1}, L_{2})_{+} - lk (L_{1}, L_{2})_{-} = (C + \frac{1}{2}) - (C - \frac{1}{2}) = 1$ . This is the stated positive-minus-negative derivative under [F1]; reversing that subtraction changes its sign.

1.2givenalgebra

A knot has one component and therefore no unordered pair of distinct components; the total pairwise linking sum contains no terms and equals $0$ .

2.1givenA1F1F2F3F4step 1.1algebra∎

If both marked crossings are mixed, write their resolution signs as $ϵ_{1}, ϵ_{2} \in {+ 1, - 1}$ and absorb every other mixed crossing into $C$ . By [F2]–[F4], $lk (L_{ϵ_{1}, ϵ_{2}}) = C + \frac{1}{2} (ϵ_{1} + ϵ_{2})$ . The alternating second difference is zero: the constant has coefficient $1 - 1 - 1 + 1 = 0$ , and each term linear in one $ϵ_{i}$ cancels when the other sign is summed. If either marked crossing is a self-crossing, the link formula [F3] is independent of that resolution, so the alternating sum over its two choices cancels term by term. Thus $D_{2} = 0$ for every pair of marked crossings by [F1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

The Fundamental Theorem of Vassiliev Invariants (standard reference, not scraped)
Introduction to Vassiliev Knot Invariants (standard reference, not scraped)