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.

Two absorbing classes retain initial-law-dependent weights

Example

Let $P = I_{2}$ . Its two states are separate absorbing closed classes. Every probability vector is stationary, and the Cesàro limit is the initial law. The global column-overlap coefficient equals one, so a global residual accuracy certificate is unavailable even though the stationary linear system is exactly solvable once an initial-law or class-weight convention is supplied.

Verification

Given: $P = I_{2}$ , an initial row law $u = (3 / 5, 2 / 5)$ , and raw column input $y = u^{T} = (3 / 5, 2 / 5)^{T}$ to the certificate algorithm.

[F1] Finite Markov closed-class graph decomposition: the algorithm returns the strongly connected components and identifies a component as closed exactly when it has no outgoing transition.

[F2] Closed-class stationary laws and initial-law-dependent Cesàro weights: all stationary laws are mixtures of the unique laws on closed classes, and initial-law weights are their first-entry probabilities.

[F3] Column-overlap formula for the zero-sum ℓ1 coefficient: $δ (P) = 1 - \min_{j, k} \sum_{i} \min (P_{i j}, P_{i k})$ .

[F4] Correct and certify a finite-chain stationary candidate: for exact rational P and a raw input y with $\sum_{i} y_{i}^{+} > 0$ , a computed coefficient at least one produces certificate-unavailable.

[F5] Correctness of finite Markov closed-class decomposition: the graph procedure returns exactly the strongly connected components and identifies as closed exactly those with no outgoing positive transition.

Proof technique: direct exact calculation.

1.1givenF1F5construct

The graph has no edge between its distinct vertices, so [F5] shows the decomposition in [F1] returns $C_{1} = {1}$ and $C_{2} = {2}$ , with no transient states. Each class law is its point mass.

2.1F2step 1.1algebra

Since $u P^{k} = u$ for every $k$ , every Cesàro average is u. By [F2], the weights are $α_{1} = u (C_{1}) = 3 / 5$ and $α_{2} = u (C_{2}) = 2 / 5$ . The full stationary family is ${(a, 1 - a)^{T} : 0 \leq a \leq 1}$ , so no unique stationary vector is selected.

3.1F3F4step 2.1algebra∎

Since $P = I_{2}$ , $P^{m} = I_{2}$ for every $m \geq 1$ . Its columns are $e_{1}$ and $e_{2}$ , whose overlap is zero, so [F3] gives $δ (P^{m}) = 1$ . The given raw input is normalized, hence [F4] returns certificate-unavailable for every $m \geq 1$ . This is consistent with the continuum of stationary laws: the strict-contraction hypothesis fails.

Depends on

Used by

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Sources

Charles M. Grinstead and J. Laurie Snell, Introduction to Probability, Chapter 11 (standard reference, not scraped)