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.

Minmod clips the slope at a smooth quadratic maximum

Example

For $u (x) = 1 - (x - h / 2)^{2}$ and cells $I_{j} = [j h, (j + 1) h)$ , the cell average in the central cell $I_{0}$ is $U_{0} = 1 - h^{2} / 12$ and its two neighbors have $U_{- 1} = U_{1} = 1 - 13 h^{2} / 12$ . The minmod slope in the central cell is zero, so the reconstruction clips the central slope at the smooth maximum.

Facts & Assumptions

Given: $h > 0$ , the three cell averages of $u (x) = 1 - (x - h / 2)^{2}$ over $I_{- 1}, I_{0}, I_{1}$ , and the MUSCL/minmod definition.

[F1]

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions: $h s_{0} = minmod (U_{0} - U_{- 1}, U_{1} - U_{0})$ and the central face states are $U_{0} \pm h s_{0} / 2$ .

Verification

1.1givenalgebra

Write $x = h / 2 + y$ on $I_{0}$ , so $y \in [- h / 2, h / 2]$ and $u = 1 - y^{2}$ . Averaging gives $U_{0} = 1 - h^{2} / 12$ . On either neighboring cell the centered coordinate $y$ has mean $\pm h$ and variance $h^{2} / 12$ , hence $U_{- 1} = U_{1} = 1 - (h^{2} + h^{2} / 12) = 1 - 13 h^{2} / 12$ .

2.1F1step 1.1algebra

Therefore $U_{0} - U_{- 1} = h^{2}$ and $U_{1} - U_{0} = - h^{2}$ . Their signs differ, so [F1] gives $s_{0} = 0$ and both reconstructed face states equal $U_{0}$ .

3.1step 1.1step 2.1algebra∎

The exact profile at either face $x = h / 2 \pm h / 2$ is $1 - h^{2} / 4$ . The reconstructed face error is $∣ U_{0} - (1 - h^{2} / 4) ∣ = h^{2} / 6$ . The reconstructed value equals the largest of the three cell averages, so this clipping introduces no new face value outside their range while retaining a second-order face error.

Depends on

Finite-volume MUSCL, nodal ENO2 and WENO5 reconstruction conventions

Used by

Nothing in the library uses this result yet.

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