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ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-09-23 (gpt-6-sol)
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.

Generic affine hyperplanes meet an embedded submanifold transversely

Example

For the unit circle S1⊆R2, the vertical line Ha={x=a} meets S1 transversely for every a∈R∖{−1,1}. For ∣a∣>1 the intersection is empty, and for ∣a∣<1 it consists of two transverse points.

Facts & Assumptions

Given: The height map h:S1→R, h(x,y)=x, and a real parameter a.

Verification

technique · direct
1.1givenalgebra

The fibre h−1(a) is S1∩Ha. When ∣a∣<1, the equation x=a on x2+y2=1 gives the two points (a,±1−a2). A tangent vector at (x,y) is (−y,x), and dh(−y,x)=−y, which is nonzero at both points because y≠0.

2.1step 1.1

For ∣a∣>1 the fibre is empty, hence transverse vacuously. For a=±1 it consists of one point with y=0, where dh vanishes and transversality fails. Thus the exceptional set is exactly the finite, hence null, set {−1,1}.

3.1step 2.1∎

Thus generic affine hyperplanes meet the embedded circle transversely.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources