ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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.

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A bounded random diffusion coefficient remains uniformly coercive

Example

On $D = (0, 1)$ , let $Y$ be uniform on $[- 1, 1]$ , take the degree cutoff $p_{1} = 0$ , set $a (x, Y) = 2 + Y$ , and set the forcing $F (Y) = 0$ . This is an affine random diffusion with explicit uniform ellipticity bounds; its unique pointwise and Bochner weak solution is zero.

Facts & Assumptions

Given: The bounded interval $D$ , uniform product law on $[- 1, 1]$ , constant coefficient terms $a_{0} = 2$ , $a_{1} = 1$ , and zero forcing.

[F1]

Finite product-law polynomial-chaos input data permits a finite product law with the stated admissible cutoff.

[F2]

Uniformly elliptic affine random diffusion model requires $0 < a_{*} \leq a (x, y) \leq a^{*}$ for every supported parameter and uses the gradient norm on $H_{0}^{1} (D)$ .

[F3]

Affine random diffusion has a measurable Bochner solution gives the unique pointwise and Bochner solution under those hypotheses.

Verification

1.1F1F2givenalgebra

At cutoff $p_{1} = 0$ , the only moment-Gram matrix is $[1]$ , so the uniform probability law is admissible by [F1]. Its support is $Γ = [- 1, 1]$ , and for every $y \in Γ$ , $1 \leq 2 + y \leq 3$ . Thus the diffusion definition applies with $a_{*} = 1$ and $a^{*} = 3$ , and $F = 0$ is a bounded affine forcing.

2.1F2F3step 1.1algebra∎

The pointwise weak equation is $\int_{0}^{1} (2 + y) u^{'} (x) v^{'} (x) d x = 0$ for every $v \in H_{0}^{1} (0, 1)$ . Taking $v = u$ and using $2 + y \geq 1$ gives $∥ u ∥_{H_{0}^{1}}^{2} \leq 0$ , so $u = 0$ for every $y$ . This solution is a strongly measurable square-integrable zero function and satisfies the Bochner equation; uniqueness also follows from [F3].

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