Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generated
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 local maximum has the wrong Morse index for the saddle certificate

Example

Let V(x,y)=−x2−y2 on U=R2. The origin is a critical point and a strict local maximum, but its Hessian has two negative eigenvalues. It therefore fails the index-one enclosure required by Validated ordered eigenvalue enclosures for an index-one Hessian and cannot be certified by Validated Hessian enclosures and a small gradient residual certify a nearby index-one saddle.

Facts & Assumptions

Given: The quadratic potential on R2 and the definitions of index-one enclosure and the local saddle certificate.

[F1]

Validated ordered eigenvalue enclosures for an index-one Hessian: an index-one enclosure requires u1≤−γ and ℓj≥γ for every j≥2.

[F2]

Validated Hessian enclosures and a small gradient residual certify a nearby index-one saddle: the nearby-saddle claim requires such an index-one enclosure at the reported point.

Given: V(x,y)=−x2−y2 and the origin.

Proof

1.1givenalgebra

The gradient is ∇V=(−2x,−2y), so it vanishes at (0,0); the Hessian is H=−2I, with ordered eigenvalues λ1=λ2=−2.

2.1F1step 1.1

For any γ>0, an interval enclosing λ2=−2 cannot have lower endpoint ℓ2≥γ>0. Thus no exact validated enclosure of this Hessian satisfies [F1].

3.1F2step 2.1∎

The point is a maximum with Morse index two, not index one. Since the required enclosure premise in [F2] fails by step 2.1, the theorem makes no certification claim for it; this example illustrates why a small gradient or critical-point report alone does not establish the desired saddle type.

Depends on

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