LemmaStatement: Literature-sourcedProof: 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.

Nonnegativity and equality for finite classical relative entropy

Statement

For probability vectors $p = (p_{i})_{i \in I}$ and $q = (q_{i})_{i \in I}$ on a finite nonempty set $I$ , define

$D (p ∥ q) : = {\begin{cases} \sum_{i : p_{i} > 0} p_{i} \ln \frac{p_{i}}{q_{i}}, & p_{i} > 0 \Rightarrow q_{i} > 0, \\ + \infty, & otherwise. \end{cases}$

Then $D (p ∥ q) \geq 0$ , and $D (p ∥ q) = 0$ if and only if $p = q$ .

Facts & Assumptions

Given: The finite set $I$ and probability vectors $p, q$ in the Statement.

[F1]

The natural logarithm as the inverse of the exponential function: $\ln x$ is the inverse exponential on $(0, \infty)$ .

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln$ is strictly increasing, continuous, and $\ln 1 = 0$ .

[F3]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $(\ln x)^{'} = 1 / x$ for $x > 0$ .

[F4]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : for a function continuous on a closed interval and differentiable inside, some interior derivative equals the endpoint slope.

Proof

1.1givenF1F2F3F4casesalgebra

Set $g (t) = t - 1 - \ln t$ for $t > 0$ . For $t > 1$ , [F3] and [F4] give $g (t) - g (1) = g^{'} (c) (t - 1)$ for some $c \in (1, t)$ , where $g^{'} (c) = 1 - 1 / c > 0$ ; for $0 < t < 1$ , they give $g (1) - g (t) = g^{'} (c) (1 - t)$ for some $c \in (t, 1)$ , where $g^{'} (c) = 1 - 1 / c < 0$ . Thus $g (t) > 0$ unless $t = 1$ , when $g (1) = 0$ . Equivalently, $- \ln t \geq 1 - t$ , with equality only at $t = 1$ .

2.1givenstep 1.1algebracases∎

If $p_{i} > 0 = q_{i}$ for some $i$ , then $D (p ∥ q) = + \infty$ by definition. Otherwise, for each $i$ with $p_{i} > 0$ , apply step 1.1 to $t = q_{i} / p_{i}$ and multiply by $p_{i}$ , obtaining $p_{i} \ln (p_{i} / q_{i}) \geq p_{i} - q_{i}$ . Summing gives $D (p ∥ q) \geq 1 - \sum_{i : p_{i} > 0} q_{i} \geq 0$ . Equality forces $q_{i} = p_{i}$ on the positive support by step 1.1 and no $q$ -mass outside it by the final inequality; hence $p = q$ . Conversely, $p = q$ makes every summand zero.

Depends on

Used by

Dependency tree · two levels

20 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

John Watrous, The Theory of Quantum Information, §5.1 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, arXiv:1106.1445v8, §§10.5 and 10.7 (standard reference, not scraped)