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.

A three-state birth–death generator and detailed balance

Example

On states ${0, 1, 2}$ , let the birth rates be $2$ from 0 to 1 and $2$ from 1 to 2, and the death rates be $1$ from 1 to 0 and $1$ from 2 to 1. The column generator, uniformization matrix at $ν = 3$ , and stationary probability are

$Q = (\begin{array}{ccc} - 2 & 1 & 0 \\ 2 & - 3 & 1 \\ 0 & 2 & - 1 \end{array}), P = I + Q / 3 = (\begin{array}{ccc} 1 / 3 & 1 / 3 & 0 \\ 2 / 3 & 0 & 1 / 3 \\ 0 & 2 / 3 & 2 / 3 \end{array}), π = (1, 2, 4)^{T} / 7.$

Detailed balance verifies the law. Moreover, $δ (P^{2}) = 2 / 9$ , so the stationary residual certificate is available, and at $x = π$ its bound is zero.

Verification

Given: The displayed column generator Q, rate $ν = 3$ , matrix P, probability vector $π$ , and raw candidate $y = π$ .

[F1] Finite column generator: off-diagonal generator entries are nonnegative, columns sum to zero, and the source column's exit rate is $r_{j} = - Q_{j j}$ .

[F2] Correct and certify a finite-chain stationary candidate: a nonnegative normalized input is unchanged by correction, so the output candidate is $x = y$ with correction distance $c = 0$ .

[F3] Uniformized finite-generator stationary residual certificate: for an admissible uniformization with $q = δ (P^{m}) < 1$ , the corrected raw candidate y satisfies $∥ y - π ∥_{1} \leq c + m ∥ Q x ∥_{1} / (ν (1 - q))$ .

[F4] Detailed balance implies invariance: for a row transition matrix K and finite state measure $μ$ satisfying $μ_{i} K_{i j} = μ_{j} K_{j i}$ , one has $μ K = μ$ ; if its mass is one, it is an invariant probability.

Proof technique: direct exact calculation; set $K = P^{T}$ .

1.1givenF1algebra

The off-diagonal entries of Q are nonnegative and its columns sum to zero; the exit rates are $2, 3, 1$ . Thus $ν = 3$ is admissible in [F1]. Directly, $P = I + Q / 3$ is the displayed nonnegative matrix and its columns each sum to one.

2.1F1F4step 1.1algebra

The coordinates of $π$ are nonnegative and sum to one. In the row kernel K, $π_{0} K_{01} = (1 / 7) (2 / 3) = 2 / 21 = π_{1} K_{10}$ and $π_{1} K_{12} = (2 / 7) (2 / 3) = 4 / 21 = π_{2} K_{21}$ ; the reverse equalities hold, all non-neighbor flows are zero in both directions, and diagonal balance is automatic. Therefore [F4] gives the row law $π^{T} K = π^{T}$ , equivalently $P π = π$ . Direct multiplication also gives $Q π = 0$ .

3.1F2F3step 2.1algebra∎

Multiplication gives $P^{2} = (\begin{array}{ccc} 1 / 3 & 1 / 9 & 1 / 9 \\ 2 / 9 & 4 / 9 & 2 / 9 \\ 4 / 9 & 4 / 9 & 2 / 3 \end{array})$ . Each pair of distinct columns has overlap $7 / 9$ , and each self-pair has overlap $1$ ; hence the minimum overlap is $7 / 9$ and the coefficient is $q = 1 - 7 / 9 = 2 / 9 < 1$ for $m = 2$ . With raw candidate $y = π$ , [F2] gives $x = π$ and $c = 0$ ; also $Q x = 0$ . Applying [F3] certifies the exact stationary law with error bound zero.

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Sources

Charles M. Grinstead and J. Laurie Snell, Introduction to Probability, Chapter 11 (standard reference, not scraped)
Yuanyuan Liu, Perturbation Bounds for the Stationary Distributions of Markov Chains (standard reference, not scraped)