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.

The factor of two between density L1 and library total variation

Example

Under the library AC convention for density representatives, on ${0, 1}$ with counting measure let $P = δ_{0}$ and $Q = δ_{1}$ . Their density $L^{1}$ distance is $2$ , while their library event-supremum total variation is $1$ and their root-L2 Hellinger distance is $\sqrt{2}$ . Thus the factor in $h^{2} \leq 2 T V$ is sharp.

Facts & Assumptions

Given: The two point-mass laws on the two-point measurable space and counting measure.

[A1]

AC is inherited from the likelihood-density convention used by the Hellinger definition; the two densities here are explicitly supplied. The Axiom of Choice

[F1]

For densities $p, q$ against a common measure, $h (P, Q)^{2} = \int (\sqrt{p} - \sqrt{q})^{2} d μ$ . Root-L2 Hellinger distance for dominated laws

[F2]

Library total variation is $\sup_{A} ∣ P (A) - Q (A) ∣$ . Total variation distance for probability laws

[F3]

For this convention, $T V (P, Q) \leq h (P, Q)$ and $h (P, Q)^{2} \leq 2 T V (P, Q)$ . Hellinger invariance and comparison with library total variation

Verification

Given: The two probability laws and their counting-measure densities.

1.1F2givenalgebra

The densities are $p (0) = 1, p (1) = 0$ and $q (0) = 0, q (1) = 1$ , so $\sum_{x \in {0, 1}} ∣ p (x) - q (x) ∣ = 2$ . The four measurable events have probability differences $0, 1, - 1, 0$ , hence the supremum in [F2] is $1$ .

2.1A1F1F3step 1.1algebra∎

By [F1], $h (P, Q)^{2} = (\sqrt{1} - \sqrt{0})^{2} + (\sqrt{0} - \sqrt{1})^{2} = 2$ , so $h (P, Q) = \sqrt{2}$ . Therefore $h (P, Q)^{2} = 2 T V (P, Q)$ , attaining equality in the upper comparison in [F3]; the lower comparison reads $1 \leq \sqrt{2}$ .

Depends on

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