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.

Three species with weighted number-to-mass conservation

Example

Let state domain and factorization neighborhood be $Ω = U = R^{3}$ , and let $K = {(2, 0, 0)}$ . Let species $A, B, C$ have molecular weights $2, 1, 1$ . In number variables let the reactions be $A ⇄ B + C$ , with forward event rate $r = 2 n_{A}$ and reverse event rate $s = n_{B} n_{C}$ . Convert to mass variables $y = (2 n_{A}, n_{B}, n_{C})$ . Verify the conservative transfer representation and one zero-compatible step from $y^{n} = (2, 0, 0)$ with $h = 1$ .

Verification

Given: The reactions and molecular weights in the Example, mass coordinates $y = (2 n_{A}, n_{B}, n_{C})$ , and step $h = 1$ .

[F1] Finite conservative production destruction systems: paired rates $p_{i j} = d_{j i}$ define a finite closed PDS and preserve the sum in its internal-transfer form.

[F2] Source-factored transfer rates for zero compartments: rates $p_{i j} = y_{j} a_{i j}$ and $d_{i j} = y_{i} a_{j i}$ have a ratio-free representation at zero coordinates.

[F3] Zero-compatible modified Patankar Euler in rate-factor form: the step solves $(I - h B (y^{n})) y^{n + 1} = y^{n}$ using donor-rate factors $a_{i j} (y^{n})$ .

[F4] The factorized zero-state step is well-defined, nonnegative, conservative, and locally first order: for the stated factorization the output is nonnegative and conserves total mass, including at zero compartments.

1.1givenF1algebra

One forward event consumes mass $2$ from $A$ and produces mass $1$ in each of $B, C$ ; one reverse event consumes mass $1$ from each of $B, C$ and produces mass $2$ in $A$ . Thus the mass transfer rates are $p_{21} = p_{31} = d_{12} = d_{13} = r = y_{1}$ and $p_{12} = p_{13} = d_{21} = d_{31} = s = y_{2} y_{3}$ , with all other rates zero. These paired rates satisfy [F1], and $y_{1} + y_{2} + y_{3} = 2 n_{A} + n_{B} + n_{C}$ is the molecular-weight mass.

1.2F2F3algebra

The transfer factors are $a_{21} = a_{31} = 1$ , $a_{12} = y_{3}$ , and $a_{13} = y_{2}$ , with all other factors zero; hence the rates have the form required by [F2], even when $y_{2}$ or $y_{3}$ is zero. At $y^{n} = (2, 0, 0)$ , the matrix in [F3] is $I - B (y^{n}) = (\begin{array}{ccc} 3 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array})$ . Solving it against $(2, 0, 0)^{T}$ gives $y^{n + 1} = (2 / 3, 2 / 3, 2 / 3)^{T}$ .

2.1F1F4step 1.1step 1.2algebra∎

The updated coordinates are nonnegative, and $2 / 3 + 2 / 3 + 2 / 3 = 2 = 2 + 0 + 0$ . This is the weighted number-to-mass conservation verified in step 1.1; it also agrees with the general zero-compatible theorem [F4].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.