CounterexampleConstruction: 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.

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An arbitrarily small residual does not certify the candidate's nonnegativity

Statement refuted

For every conservative Patankar linear system, a sufficiently small linear residual by itself certifies that the computed candidate is nonnegative.

Witness

Take state domain $Ω = R$ and one compartment with no transfers, so its Patankar matrix is $M = (1)$ and its nonnegative right-hand side is $b = 0$ . For any $ε > 0$ , the candidate $\hat{y} = - ε$ has residual $M \hat{y} - b = - ε$ , which can be arbitrarily small while the candidate is negative.

Facts & Assumptions

Given: A one-compartment closed system with no transfers, right-hand side $0$ , and any real $ε > 0$ .

[F1]

Finite conservative production destruction systems permits $m = 1$ and zero paired transfer rates.

[F2]

Modified Patankar Euler for conservative transfer systems with any positive denominator and zero rates has solve matrix $M = (1)$ .

[F3]

Certified residuals bound mass drift and the exact-solve positivity margin: under its stored-operand and successful-arithmetic hypotheses, the componentwise margin is sufficient to certify nonnegativity of the exact solution; the residual alone is not a sign condition on the candidate.

Refutation

1.1givenF1F2algebra

Set $m = 1$ , $p_{11} = d_{11} = 0$ , $h = 1$ , $y^{n} = 0$ , and $σ_{1} = 1$ . These data satisfy [F1] and the admissible input conditions of [F2], and the system $M y^{+} = y^{n}$ is $y^{+} = 0$ .

2.1step 1.1algebrachoose

For the candidate $\hat{y} = - ε$ , the exact residual is $r = M \hat{y} - b = - ε$ , so $∣ r ∣ = ε$ can be made smaller than any prescribed positive tolerance while $\hat{y} < 0$ .

3.1F3step 2.1algebra∎

Here the exact solve is $y^{*} = 0$ . Whenever the stored-operand and successful-arithmetic hypotheses of [F3] hold, its radius satisfies $ρ \geq ε$ , so its sufficient margin $\hat{y} \geq ρ$ fails. For every real $ε > 0$ , step 2.1 already gives a negative candidate with exact residual magnitude $ε$ , refuting the statement independently of floating-point evaluation.

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