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.

Finite credit reserve: certified absorption statistics

Example

A credit reserve has levels $0, 1, 2, 3, 4$ . From an interior level it moves one level up or down with equal probability; levels 0 and 4 are absorbing failure and success states. For transient levels $U = {1, 2, 3}$ , the transient block is

$T = (\begin{array}{ccc} 0 & 1 / 2 & 0 \\ 1 / 2 & 0 & 1 / 2 \\ 0 & 1 / 2 & 0 \end{array}), b_{4} = (0, 0, 1 / 2)^{T} .$

The chance of eventual success from level i is $i / 4$ and the mean number of steps to absorption is $i (4 - i)$ . Starting from level 2, the two absorbing class weights are both $1 / 2$ . A deliberately perturbed numerical solve illustrates the residual certificate.

Verification

Given: The five-state transition rule above, transient set U, and success state 4.

[F1] Transient hitting, mean exit, and residual bounds: if the finite transient block has the escape path property, then $h = (I - T)^{- 1} b_{4}$ , $t = (I - T)^{- 1} 1$ , and both statistics have residual error bounded by $∥ (I - T)^{- 1} ∥_{\infty}$ times the equation residual.

[F2] Closed-class stationary laws and initial-law-dependent Cesàro weights: eventual class weights depend on the initial law and are computed by the transient-block exit system.

Proof technique: direct exact calculation.

1.1givenF1construct

From level 1, the path down exits within one step with probability $1 / 2$ ; from level 2, the two paths down-down and up-up exit within two steps with total probability $1 / 2$ ; from level 3, the path up exits within one step with probability $1 / 2$ . Thus every transient state escapes within $L = 2$ with probability at least $ε = 1 / 2$ , and [F1] gives $∥ (I - T)^{- 1} ∥_{\infty} \leq 4$ .

1.2F1givenalgebra

Let $\hat{h} = (13 / 50, 1 / 2, 37 / 50)^{T}$ . Direct multiplication gives $b_{4} - (I - T) \hat{h} = (- 1 / 100, 0, 1 / 100)^{T}$ , with infinity norm $1 / 100$ .

2.1F1step 1.1algebra

The success equations are $h_{1} = h_{2} / 2$ , $h_{2} = (h_{1} + h_{3}) / 2$ , and $h_{3} = 1 / 2 + h_{2} / 2$ . Substitution gives $h = (1 / 4, 1 / 2, 3 / 4)^{T}$ , which satisfies $(I - T) h = b_{4}$ by direct multiplication. The mean-exit equations are $t_{1} = 1 + t_{2} / 2$ , $t_{2} = 1 + (t_{1} + t_{3}) / 2$ , and $t_{3} = 1 + t_{2} / 2$ ; direct substitution verifies $t = (3, 4, 3)^{T}$ .

3.1F1F2step 1.1step 2.1step 1.2algebra∎

By [F1] and step 1.1, $∥ h - \hat{h} ∥_{\infty} \leq 4 / 100 = 1 / 25$ . The exact error is $1 / 100$ , so the finite-step escape certificate is valid and conservative. From level 2, step 2.1 gives success weight $1 / 2$ ; because the two closed classes exhaust probability one, the failure weight is also $1 / 2$ , as in [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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.

Sources

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