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.

An unanchored component is rejected before solving

Example

Let $V = {1, 2, 3}$ , with the single unit-weight edge ${1, 2}$ , labeled set $B = {1}$ , boundary value $y_{1} = 0$ , and unlabeled set $U = {2, 3}$ . The isolated component ${3}$ contains no labeled vertex, so the graph-harmonic propagation algorithm returns status $unanchored-component$ before forming a linear solve.

Graph diagram

The edge between 1 and 2 has weight 1; vertex 1 has boundary value 0. Vertices 2 and 3 are unlabeled, and vertex 3 is isolated. The verification checks the support components directly.

Facts & Assumptions

Given: Three vertices, one unit-weight edge ${1, 2}$ , $B = {1}$ , $y_{1} = 0$ , and $U = {2, 3}$ .

[A1]

$w_{12} = w_{21} = 1$ and every other weight is zero.

[F1]

Graph-harmonic label extension on a finite weighted graph: anchoring requires every support component meeting $U$ to meet $B$ .

[F2]

Finite graph-harmonic label propagation with a residual status: the algorithm checks component anchors and returns $unanchored-component$ before constructing the SPD solve if any component meeting $U$ misses $B$ .

Verification

1.1givenA1F1algebra

The support components are ${1, 2}$ and ${3}$ . The first meets both $B$ and $U$ , while ${3} \subseteq U$ and ${3} \cap B = \emptyset$ , so the anchor condition in [F1] fails.

2.1givenF2step 1.1∎

By [F2], this failed anchor check returns status $unanchored-component$ before forming a linear system; no singular solve is attempted.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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