Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + 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.

The boundary cycle of a round annulus has index 1 inside the annulus and 0 on either side

Example

Let p∈C and 0<r1<r2, and let Cj(t)=p+rjexp⁡(it) on [0,2π] for j=1,2. Let Γ be the complex chain ((1,C2),(−1,C1)), written C2−C1. Then Γ is a cycle with trace {∣z−p∣=r1}∪{∣z−p∣=r2}, and

n(Γ,z)={0,∣z−p∣<r1,1,r1<∣z−p∣<r2,0,∣z−p∣>r2.

Let s1,s2 be reals with 0<s1<r1 and r2<s2, and put Ω={z:s1<∣z−p∣<s2}. Then Γ has trace in the open set Ω and is null-homologous in Ω. The ambient open set is named before the homology because the notion depends on it: the smaller annulus {r1<∣z−p∣<r2} does not contain the trace of Γ and is therefore not an open set in which Γ is a chain at all.

Facts & Assumptions

Given: A point p, radii 0<r1<r2, the circles C1,C2 above, and the chain Γ=C2−C1.

[L1]

The trace of a sum of chains is the union of their traces, the negative of a chain has the same trace, a sum of cycles and the negative of a cycle are cycles, and for p′ off the traces involved n(Γ1+Γ2,p′)=n(Γ1,p′)+n(Γ2,p′) and n(−Γ,p′)=−n(Γ,p′) (Chain integration and the index are additive in the chain, and reverse with it).

[L2]

For a∈C, r>0 and k∈Z, the contour γk(t)=a+rexp⁡(ikt) on [0,2π] is a closed complex contour with n(γk,z)=k for ∣z−a∣<r and n(γk,z)=0 for ∣z−a∣>r; for k≠0 its trace is {z:∣z−a∣=r} (A circle traversed k times has winding number k inside and 0 outside).

[L3]

A complex chain is a finite list of pairs (mk,γk); a list of closed contours is a cycle; the negative of a chain negates every coefficient; and a single closed contour with coefficient 1 is a cycle whose trace is that contour's trace (Complex chains, their traces, and cycles).

[L4]

n(Γ,p′)=(2πi)−1∫Γdz/(z−p′), and for a single closed contour with coefficient 1 this is that contour's winding number (Integration over a complex chain and the index of a chain).

[L5]

A cycle Γ with trace in an open Ω is null-homologous in Ω when n(Γ,q)=0 for every q∈C∖Ω (Null-homologous cycles and homologous cycles in an open set).

Verification

technique · direct
1.1givenL2

By [L2] with k=1 each Cj is a closed complex contour with trace {∣z−p∣=rj}, with n(Cj,z)=1 for ∣z−p∣<rj and n(Cj,z)=0 for ∣z−p∣>rj.

2.1step 1.1L1L3L4

By [L1] and [L3] the chain Γ=C2−C1 is a cycle and its trace is {∣z−p∣=r1}∪{∣z−p∣=r2}, and by [L1] and [L4] its index off that trace is n(C2,z)−n(C1,z).

3.1step 1.1step 2.1

Evaluating step 2.1 with step 1.1: for ∣z−p∣<r1<r2 the value is 1−1=0; for r1<∣z−p∣<r2 it is 1−0=1; for ∣z−p∣>r2>r1 it is 0−0=0.

4.1step 2.1step 3.1L5∎

With 0<s1<r1 and r2<s2 the trace of Γ lies in Ω={s1<∣z−p∣<s2}, and C∖Ω={∣z−p∣≤s1}∪{∣z−p∣≥s2}; every point of the first set has ∣z−p∣≤s1<r1 and every point of the second has ∣z−p∣≥s2>r2, so step 3.1 gives index 0 at each, and [L5] makes Γ null-homologous in Ω.

Depends on

Used by

Dependency tree · two levels

49 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