Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-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.

Worked case: a normal disk in the zero-crossing unknot exterior

Example

For the crossingless oriented unknot code, triangulate its solid-torus exterior D2×S1 by triangulating D2 as one triangle with vertices a0,a1,a2 and S1 as the three-edge cycle b0b1b2b0. In the prism over b0b1, write ai0=(ai,b0) and ai1=(ai,b1) and use the tetrahedra T1=[a00,a10,a20,a21],T2=[a00,a10,a11,a21],T3=[a00,a01,a11,a21]. The properly embedded normal disk D2×{s∗} at the midpoint s∗ of b0b1 meets T1 in the triangle cutting off a21, T2 in the quadrilateral separating {a00,a10} from {a11,a21}, and T3 in the triangle cutting off a00; every other normal coordinate is zero. It is connected, has nonempty boundary and Euler characteristic one, and its boundary meets the marked original meridian μ={a0}×S1 once mod two.

\begin{tikzpicture}[scale=1]
\coordinate (A0) at (0,0); \coordinate (A1) at (2,0); \coordinate (A2) at (1,1.5);
\coordinate (B0) at (0.6,0.7); \coordinate (B1) at (2.6,0.7); \coordinate (B2) at (1.6,2.2);
\draw (A0)--(A1)--(A2)--cycle; \draw (B0)--(B1)--(B2)--cycle;
\draw (A0)--(B0) (A1)--(B1) (A2)--(B2);
\draw[dashed] (A0)--(B1) (A0)--(B2) (A1)--(B2);
\coordinate (M0) at (0.3,0.35); \coordinate (M1) at (2.3,0.35); \coordinate (M2) at (1.3,1.85);
\draw[red,very thick] (M0)--(M1)--(M2)--cycle;
\node[below left] at (A0) {$a_0^0$}; \node[below right] at (A1) {$a_1^0$}; \node[left] at (A2) {$a_2^0$};
\node[left] at (B0) {$a_0^1$}; \node[right] at (B1) {$a_1^1$}; \node[above] at (B2) {$a_2^1$};
\node[red] at (1.35,0.8) {$D^2\times\{s_*\}$};
\end{tikzpicture}

The dashed edges indicate the staircase prism subdivision specified by T1,T2,T3; the red section is the geometric disk whose intersections with those tetrahedra give the stated triangle–quadrilateral–triangle vector. The formula for the tetrahedra, not the schematic projection, fixes the exact triangulation.

Example references

Normal-surface coordinates and the Haken cone — triangle and quadrilateral normal disk types.

Linear-size good exterior triangulation from a planar knot code — fixed crossingless solid-torus exterior model and marked meridian path.

Hass–Lagarias–Pippenger, §4 and §5 — unknot exterior and normal disk coordinates.

Facts & Assumptions

Given: The product triangulation D2×S1 of the exterior and its marked meridian path.

[F1]

Normal-surface coordinates and the Haken cone: a triangle or quadrilateral normal disk meets tetrahedral faces in matching normal arcs; admissibility allows at most one quadrilateral type per tetrahedron.

[F2]

Linear-size good exterior triangulation from a planar knot code: at zero crossings the exterior is a fixed triangulated solid torus with the original meridian represented by a three-edge fibre cycle.

Verification

1.1givenF2construct

For the prism over b0b1, write ai0=(ai,b0) and ai1=(ai,b1). Use the staircase tetrahedra T1=[a00,a10,a20,a21],T2=[a00,a10,a11,a21],T3=[a00,a01,a11,a21]. Their common faces are triangulated by the same ordered diagonals, so the three tetrahedra form the triangular-prism block; repeating this order on the three-cycle of S1 gives the product triangulation in [F2].

2.1F1step 1.1construct

The level disk at the midpoint of the product edge separates the bottom vertices ai0 from the top vertices ai1. In T1 it cuts off the single top vertex a21, giving one triangle; in T2 it separates the pair {a00,a10} from {a11,a21}, giving one quadrilateral; in T3 it cuts off the single bottom vertex a00, giving one triangle. It misses all tetrahedra in the other two prism blocks. Its intersections on shared faces agree because they are the same level section of the product, so the matching equations hold. Only one quadrilateral coordinate is nonzero, so [F1] gives an admissible normal disk vector.

3.1givenF2step 2.1construct

The geometric surface is exactly D2×{s∗}, hence is a connected disk. Its boundary is the nonempty circle ∂D2×{s∗}. The disk itself has Euler characteristic 1. The marked meridian {a0}×S1 passes through exactly one point of this boundary at (a0,s∗), in the interior of the chosen fiber edge; the intersection is transverse and has parity 1.

4.1F1F2step 2.1step 3.1algebra∎

Thus the explicit vector satisfies the matching, quadrilateral, connectedness, boundary, Euler-characteristic and mod-two meridian checks claimed in the statement.

Depends on

Used by

Nothing in the library uses this result yet.

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