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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The diagram linking-number scan terminates and returns the stated integer

Statement

Assume the Axiom of Countable Choice. On every finite regular oriented diagram with exactly two components, the crossing-scan algorithm halts after one iteration per crossing and returns the integer half-sum of the mixed crossing signs. On an input with any other number of components it returns invalid-input after the finite component-count check.

Facts & Assumptions

Given: The Axiom of Countable Choice and a finite regular oriented diagram D with its declared component set.

[A1]

The Axiom of Countable Choice, written ACω (The Axiom of Countable Choice (ACω)).

[F1]

A regular oriented diagram has finitely many transverse double points and finitely many link components (Regular oriented link diagrams).

[F2]

On the two-component input the procedure scans each crossing once, adds its sign exactly when its branches lie on distinct components, and returns half the resulting sum (Compute oriented linking number from a diagram).

[F3]

For every regular two-component diagram, the mixed crossing sum is even and its half is the integer linking number (The mixed crossing sum is even and defines linking number).

Proof

1.1A1F1F2givenalgebra

The component-count check terminates because the input declares a finite component set. It returns invalid-input when that cardinality is not two. Otherwise the finite crossing list has some length N∈N0 by [F1], and the algorithm performs exactly one update for each of those N crossings with no recursive call or unbounded search. Hence the scan halts after exactly N iterations.

2.1A1F2F3step 1.1∎

In the accepted case, the update in [F2] adds precisely each summand of S(D) and no self-crossing. Thus the returned value is S(D)/2; [F3] shows this is an integer and is the linking number specified in the lemma. All arithmetic is a finite integer sum.

Depends on

Used by

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Sources