ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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 two-response Gibbs policy with exact probabilities

Example

Take $Y = {a, b}$ , reference probabilities $q_{a} = q_{b} = 1 / 2$ , any $β > 0$ , and rewards $r (a) = 0$ and $r (b) = β \log 3$ . The partition sum is $Z = 2$ , the unique maximizing policy is $p^{*} (a) = 1 / 4$ , $p^{*} (b) = 3 / 4$ , and the maximum objective value is $F_{q, r, β} (p^{*}) = β \log 2$ .

Example references

Finite KL-regularized policy improvement: defines the objective and partition inputs.

Finite-action Gibbs policy theorem: gives the maximizing policy and objective-gap identity.

Natural logarithm: ensures $e^{\log 3} = 3$ .

Facts & Assumptions

Given: $Y = {a, b}$ , $q = (1 / 2, 1 / 2)$ , $β > 0$ , and $r (a) = 0$ , $r (b) = β \log 3$ .

[F1]

Finite KL-regularized policy improvement: $F_{q, r, β} (p) = \sum_{y} p_{y} r (y) - β D (p ∥ q)$ on the finite probability simplex.

[F2]

The finite-action Gibbs policy uniquely maximizes reward minus KL: $Z = \sum_{y} q_{y} e^{r (y) / β}$ , $p_{y}^{*} = q_{y} e^{r (y) / β} / Z$ , and $F (p^{*}) - F (p) = β D (p ∥ p^{*})$ , with equality only at $p = p^{*}$ .

[F3]

The natural logarithm as the inverse of the exponential function: $e^{\log t} = t$ for every $t > 0$ .

Verification

1.1F2F3givenalgebra

By [F3], $e^{r (a) / β} = 1$ and $e^{r (b) / β} = e^{\log 3} = 3$ . Hence $Z = (1 / 2) 1 + (1 / 2) 3 = 2$ , and [F2] gives $p^{*} (a) = 1 / 4$ and $p^{*} (b) = 3 / 4$ .

2.1F1F2givenstep 1.1algebra∎

By [F1], the objective being maximized is $F_{q, r, β}$ . Its gap identity in [F2] gives $F (p^{*}) - F (p) = β D (p ∥ p^{*}) \geq 0$ for every probability vector $p$ , with equality only at $p = p^{*}$ . Substituting $Z = 2$ into $F (p^{*}) = β \log Z$ gives the maximum value $β \log 2$ .

Depends on

Used by

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