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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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 scalar ridge penalty and its validation hypergradient

Example

Let the hyperparameter be λ∈(0,∞) and the inner state be w∈R. For one unit-feature, unit-response training datum, define the scalar ridge objective f(λ,w)=12(w−1)2+λ2w2, and define the validation objective g(w,λ)=12w2. The unique inner minimizer is w∗(λ)=1/(1+λ), so J(λ)=g(w∗(λ),λ)=1/(2(1+λ)2) and J′(λ)=−1/(1+λ)3. The local hypergradient formula gives the same derivative.

Facts & Assumptions

Given: λ>0, f(λ,w)=12(w−1)2+λ2w2, and g(w,λ)=12w2. Verify the stationary branch, direct outer derivative, and hypergradient.

[F1]

Smooth bilevel stationary problems and their hypergradients: H=∂w∇wf, B=∂λ∇wf, J(λ)=g(w∗(λ),λ), and gradients are column vectors.

[F2]

Hypergradient of a smooth local stationary branch: for an invertible stationary Hessian, J′=∂λg−BTH−T∇wg.

Verification

1.1givenF1algebra

Differentiate the inner loss: ∂wf=(1+λ)w−1 and H=1+λ>0. Thus its stationary point is w∗=1/(1+λ). Completing the square gives f(λ,w)=12(1+λ)(w−11+λ)2+λ2(1+λ), so this stationary point is the unique global minimizer.

2.1givenstep 1.1algebra

Substitution into the outer loss gives J(λ)=1/(2(1+λ)2), whose direct derivative is J′(λ)=−1/(1+λ)3.

3.1F1F2step 1.1step 2.1algebra∎

At w∗, the mixed derivative is B=∂λ((1+λ)w−1)=w∗=1/(1+λ), the state derivative is ∇wg=w∗=1/(1+λ), and ∂λg=0. Formula [F2] therefore gives −BH−1∇wg=−1/(1+λ)3, agreeing with step 2.1.

Depends on

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Dependency tree · two levels

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