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 coefficient jump creates nonzero edge-flux oscillation

Example

Assume Countable Choice for the cited Sobolev/L2 interface.

Let $E = {(0, y) : 0 \leq y \leq 1}$ be shared by $K^{-} = conv {(- 1, 0), (0, 0), (0, 1)}$ and $K^{+} = conv {(1, 0), (0, 0), (0, 1)}$ . Set $A^{-} = 1$ , $A^{+} = 1 + y$ , and $U (x, y) = x$ . On each triangle $A$ is positive and Lipschitz; its traces differ on $E$ .

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, The piecewise-Lipschitz coefficient interface and projected edge flux oscillation of Residual indicators and data oscillation for P1 AFEM.

[F1]

For a scalar diffusion coefficient, the outward conormal flux on a side is $A \nabla U \cdot n$ ; edge oscillation is the squared $L^{2}$ distance of the outward flux sum from its edge mean (Residual indicators and data oscillation for P1 AFEM).

Verification

1.1A1F1givenalgebra

The gradient is $\nabla U = (1, 0)$ . The outward normals on $E$ are $n^{-} = (1, 0)$ and $n^{+} = (- 1, 0)$ , so the sum of outward conormal fluxes is $J_{E} (y) = 1 - (1 + y) = - y$ . Its edge mean on $[0, 1]$ is $- 1 / 2$ .

2.1step 1.1algebra∎

Since $h_{E} = 1$ , $h_{E} ∥ J_{E} - Π_{E}^{0} J_{E} ∥_{L^{2} (E)}^{2} = \int_{0}^{1} (y - \frac{1}{2})^{2} d y = \frac{1}{12} > 0.$ Thus the coefficient interface creates a nonzero projected edge contribution. No general reliability or complexity assertion for this coefficient model follows from the example.

Finite check

The edge integral is the elementary variance of $y$ on $[0, 1]$ , equal to $1 / 12$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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 2 (standard reference, not scraped)
Bonito, Canuto, Nochetto & Veeser, Adaptive Finite Element Methods (2024) (standard reference, not scraped)