LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Finite Gibbs relative entropy and free-energy minimum

Statement

Let $S$ be a finite nonempty set, $H : S \to R$ , and $β > 0$ . Define $Z = \sum_{s \in S} e^{- β H (s)}$ and $γ (s) = Z^{- 1} e^{- β H (s)}$ . For every probability vector $p$ on $S$ , with $0 \log 0 = 0$ , its Shannon entropy $H_{S h} (p) = - \sum_{s} p (s) \log p (s)$ and relative entropy $D (p ∥ γ) = \sum_{p (s) > 0} p (s) \log (p (s) / γ (s))$ satisfy $D (p ∥ γ) = - H_{S h} (p) + β \sum_{s} p (s) H (s) + \log Z \geq 0.$ Equality holds exactly when $p = γ$ . Equivalently, $γ$ is the unique minimizer of $\sum_{s} p (s) H (s) - β^{- 1} H_{S h} (p)$ , and the minimum is $- β^{- 1} \log Z$ .

Facts & Assumptions

Given: A finite nonempty state set $S$ , a real-valued energy $H$ , an inverse-temperature parameter $β > 0$ , and a probability vector $p$ .

[A1]

Every term $e^{- β H (s)}$ is a positive finite real number; hence $0 < Z < \infty$ and $γ$ is a strictly positive probability vector.

[F1]

For $r > 0$ , $\log r = \int_{1}^{r} d t / t$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Proof

1.1A1F1algebra

For every $r > 0$ , [F1] gives $- \log r = \int_{r}^{1} d t / t \geq 1 - r$ when $r < 1$ , and $- \log r = - \int_{1}^{r} d t / t \geq 1 - r$ when $r > 1$ ; both comparisons are strict unless $r = 1$ . For $p (s) > 0$ , put $r_{s} = γ (s) / p (s) > 0$ . Then $p (s) \log (p (s) / γ (s)) \geq p (s) - γ (s)$ . Summing over the support of $p$ yields $D (p ∥ γ) \geq 1 - \sum_{p (s) > 0} γ (s) \geq 0$ .

2.1A1step 1.1algebra

If $D (p ∥ γ) = 0$ , the two nonnegative gaps in step 1.1 both vanish: every scalar inequality on the support is equality and the omitted $γ$ -mass is zero. The strictness just proved gives $γ (s) / p (s) = 1$ for every $p (s) > 0$ , while positivity of $γ$ makes zero omitted mass imply $p (s) > 0$ for every $s$ . Thus $p = γ$ . Conversely $p = γ$ makes every summand zero.

3.1step 1.1step 2.1algebra∎

Expanding $\log γ (s) = - β H (s) - \log Z$ in the finite sum defining $D$ gives the displayed identity. Rearranging the nonnegativity and equality statement from steps 1.1–2.1 proves the unique-minimum formulation.

Depends on

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t

Finite Gibbs relative entropy and free-energy minimum

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources