ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-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.

A coarse mesh can miss the convection smallness condition

Example

Assume Countable Choice for the cited Sobolev/L2 interface.

Fix a positive mesh diameter $h_{0}$ , exponent $s > 0$ , and a stipulated positive dual-pickup constant $C_{D}$ . The sufficient small-mesh parameter in Quasi-orthogonality for constant convection on a fine initial mesh is $μ = C_{D} h_{0}^{s} ∣ b ∣$ .

Facts & Assumptions

[A1]

Countable Choice is the background assumption in the cited Sobolev and $L^{2}$ suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Given: Countable Choice, A prescribed mesh scale $h_{0} > 0$ , $s > 0$ , $C_{D} > 0$ , and a constant divergence-free vector $b \in R^{2}$ .

[F1]

The nonsymmetric quasi-orthogonality estimate assumes $μ = C_{D} h_{0}^{s} ∣ b ∣ < 1$ (Quasi-orthogonality for constant convection on a fine initial mesh).

Verification

1.1A1F1givenconstructalgebra

Choose any unit vector $e$ and set $b = 2 (C_{D} h_{0}^{s})^{- 1} e$ . This is constant and divergence-free, and direct substitution gives $μ = 2$ .

2.1step 1.1algebra∎

The strict sufficient condition $μ < 1$ therefore fails on this prescribed mesh. This only shows that the stated sufficient estimate cannot be invoked from its hypotheses; it does not imply failure of convergence or of the PDE solution.

Finite check

The substitution is exact: $C_{D} h_{0}^{s} \cdot 2 (C_{D} h_{0}^{s})^{- 1} = 2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Ricardo H. Nochetto, Adaptive Finite Element Methods, Lecture 3 (standard reference, not scraped)
Bonito, Canuto, Nochetto & Veeser, Adaptive Finite Element Methods (2024) (standard reference, not scraped)