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-function loss class has explicit Rademacher complexity

Example

Let $Z = {a, b}$ with the discrete sigma-algebra, let $P ({a}) = p$ and $P ({b}) = 1 - p$ for $p \in [0, 1]$ , and let $G = {g_{0}, g_{1}}$ where $g_{0}$ is identically zero and $g_{1} (z) = 1$ for $z = a$ and $g_{1} (z) = 0$ for $z = b$ . For a one-observation sample ( $m = 1$ ), ${\hat{R}}_{S} (G) = 1 / 2$ when $Z_{1} = a$ and $0$ when $Z_{1} = b$ , so $R_{1} (G) = p / 2$ .

Example references

One-sided Rademacher complexity: defines the empirical sign supremum and its expectation over the sample.

Facts & Assumptions

Given: The two-point probability law, the finite class $G$ , and sample size $m = 1$ .

[F1]

${\hat{R}}_{S} (G) = E_{σ} \sup_{g \in G} m^{- 1} \sum_{i} σ_{i} g (Z_{i})$ and $R_{m} (G) = E_{S} {\hat{R}}_{S} (G)$ . One-sided Rademacher complexity

Verification

Given: The objects and hypotheses in the Example.

1.1F1givenalgebra

If $Z_{1} = a$ , then for sign $σ_{1} = 1$ the supremum of $σ_{1} g (Z_{1})$ over $G$ is $1$ , and for $σ_{1} = - 1$ it is $0$ . Thus ${\hat{R}}_{S} (G) = \frac{1}{2} (1 + 0) = \frac{1}{2}$ . If $Z_{1} = b$ , both functions vanish at the observation, so ${\hat{R}}_{S} (G) = 0$ .

2.1F1step 1.1algebra∎

The two sample outcomes have probabilities $p$ and $1 - p$ . Averaging the values from step 1.1 gives $R_{1} (G) = p (\frac{1}{2}) + (1 - p) 0 = p / 2$ .

Depends on

One-sided Rademacher complexity

Used by

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Dependency tree · two levels

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