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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Inner quadrature error enters an outer gradient certificate

Example

The gradient of an integral objective may itself be computed by quadrature. Its quadrature discrepancy must be included separately from the computed-gradient residual.

Facts & Assumptions

Given: $f (x) = \int_{0}^{1} (x - s)^{4} d s$ on $R$ , candidate $\tilde{x} = 3 / 5$ , and the two-panel composite midpoint rule applied to the integral of $\partial_{x} (x - s)^{4}$ .

[F1]

An inexact gradient residual certifies distance under strong convexity: if the computed gradient residual is at most $η$ and its discrepancy from the true gradient is at most $δ$ , then the distance to a $μ$ -strongly convex minimizer is at most $(η + δ) / μ$ .

[F2]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the gradient is the first derivative in the standard real coordinate.

Verification

1.1F2givenalgebra

Integrating gives $f (x) = (x^{5} - (x - 1)^{5}) / 5$ , hence $f^{'} (x) = x^{4} - (x - 1)^{4}$ and $f^{''} (x) = 12 (x - 1 / 2)^{2} + 1 \geq 1$ . Thus $μ = 1$ is valid and the minimizer is $x^{*} = 1 / 2$ .

1.2F2givenalgebra

At $\tilde{x} = 3 / 5$ , the true gradient is $f^{'} (3 / 5) = 13 / 125 = 104 / 1000$ . The two midpoint nodes are $1 / 4$ and $3 / 4$ , so the composite midpoint estimate is $\hat{g} = \frac{1}{2} (4 (7 / 20)^{3} + 4 (- 3 / 20)^{3}) = 79 / 1000$ . Therefore the computed residual is $η = 79 / 1000$ and the exact discrepancy is $δ = ∣ 104 / 1000 - 79 / 1000 ∣ = 1 / 40 = 25 / 1000$ .

2.1F1step 1.1step 1.2algebra∎

The actual distance is $∣ 3 / 5 - 1 / 2 ∣ = 1 / 10 = 100 / 1000$ . By [F1] the certificate is $(η + δ) / μ = 104 / 1000$ , which is valid and close to the actual distance. The objective's inner integral was evaluated approximately, so its gradient discrepancy was not inferred from the residual.

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