ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Harmonic extension on a three-vertex path

Example

Take the path with vertices $1, 2, 3$ , unit weights on edges ${1, 2}$ and ${2, 3}$ , labeled boundary $B = {1, 3}$ , and values $y_{1} = 0$ , $y_{3} = 1$ . The unique graph-harmonic extension assigns value $1 / 2$ to vertex 2 and has Dirichlet energy $1 / 4$ .

Graph diagram

The labels inside the end vertices are their prescribed boundary values. Each drawn undirected edge has weight 1; vertex 2 is unlabeled. The calculation in Verification uses the energy formula, not the drawing.

Facts & Assumptions

Given: A three-vertex path with unit edge weights, boundary vertices $1$ and $3$ with values 0 and 1, and unlabeled vertex 2.

[A1]

$w_{12} = w_{21} = w_{23} = w_{32} = 1$ and all other weights are zero.

[F1]

Graph-harmonic label extension on a finite weighted graph: the extension minimizes $E (f) = \frac{1}{2} f^{T} L f$ over functions with the specified boundary values.

[F2]

Anchored graph-harmonic extensions are unique and obey a maximum principle: for an anchored finite graph, the minimizer is unique and solves $f_{U} = - L_{U U}^{- 1} L_{U B} y_{B}$ .

Verification

1.1givenA1F2algebra

Here $U = {2}$ , $L_{U U} = [2]$ , $L_{U B} = [- 1 - 1]$ , and $y_{B} = (0, 1)^{T}$ . By [F2], $f_{2} = - [2]^{- 1} ((- 1) 0 + (- 1) 1) = 1 / 2$ .

2.1givenA1F1F2step 1.1algebra∎

For any candidate value $t = f_{2}$ , the two edge contributions give $E (t) = \frac{1}{2} (t^{2} + (1 - t)^{2}) = \frac{1}{4} + (t - \frac{1}{2})^{2}$ . This is uniquely minimized at $t = 1 / 2$ with energy $1 / 4$ , confirming [F1] and [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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.