ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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 zero compartment makes the standard ratio undefined

Example

Take state domain and factorization neighborhood $Ω = U = R^{2}$ and $K = {(1, 0)}$ . Consider two compartments $X, Y$ with transfers $X \to Y$ at rate $2 y_{X}$ and $Y \to X$ at rate $3 y_{Y}$ , starting from $(y_{X}^{n}, y_{Y}^{n}) = (1, 0)$ . Verify that the standard denominator choice $σ_{i} = y_{i}^{n}$ is undefined, while the source-factored zero-compatible step with $h = 1$ is well-defined and conserves mass.

Verification

Given: The two transfer rates, initial state $(1, 0)$ , and step $h = 1$ .

[F1] Finite conservative production destruction systems: a transfer from donor $j$ to recipient $i$ is $p_{i j} = d_{j i}$ .

[F2] Modified Patankar Euler for conservative transfer systems: the standard step requires every declared denominator $σ_{j} > 0$ and rejects nonpositive denominators before rate evaluation; its standard consistency choice is $σ_{j} = y_{j}^{n}$ .

[F3] Source-factored transfer rates for zero compartments: a transfer from donor $j$ factors as $p_{i j} (y) = y_{j} a_{i j} (y)$ , with the coefficient defined also at zero.

[F4] Zero-compatible modified Patankar Euler in rate-factor form: the factorized update solves $(I - h B (y^{n})) y^{n + 1} = y^{n}$ without dividing by $y_{j}^{n}$ .

[F5] The factorized zero-state step is well-defined, nonnegative, conservative, and locally first order: under the source-factorization hypotheses on an open neighborhood of compact $K \subseteq [0, \infty)^{m}$ , the update is uniquely defined, nonnegative, and mass-conservative for $y^{n} \in K$ and $h > 0$ .

1.1F1F2algebra

The rates are $p_{21} = d_{12} = 2 y_{X}$ and $p_{12} = d_{21} = 3 y_{Y}$ , so [F1] holds. At $y^{n} = (1, 0)$ , the standard choice has $σ_{Y} = 0$ , so [F2] returns inadmissible-input before evaluating a rate. The displayed standard-step equation itself would contain $p_{12} (y^{n}) y_{Y}^{n + 1} / σ_{Y} = 0 \cdot y_{Y}^{n + 1} / 0$ , which is undefined; assigning $0 / 0 = 0$ is not an algebraic evaluation of that expression.

1.2F3F4F5algebra

The transfer factors $a_{21} = 2$ and $a_{12} = 3$ are nonnegative and $C^{2}$ on $U = R^{2}$ , and $y^{n} = (1, 0) \in K$ with $h = 1 > 0$ , so the hypotheses of [F5] hold. With $h = 1$ , [F4] gives $(\begin{array}{cc} 3 & - 3 \\ - 2 & 4 \end{array}) y^{n + 1} = (\begin{array}{c} 1 \\ 0 \end{array})$ , whose determinant is $6$ and whose solution is $(y_{X}^{n + 1}, y_{Y}^{n + 1}) = (2 / 3, 1 / 3)$ .

2.1F5step 1.2algebra∎

The computed coordinates are nonnegative and sum to $1$ , equal to the initial total. This agrees with [F5] and exhibits a well-defined first-order zero-state extension; it makes no second-order claim for a zero MPRK stage.

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