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.

Positive endpoint samples do not ensure a positive interpolant

Example

The quadratic $p (t) = t^{2} - t + 1 / 10$ is positive at both sampled endpoints $t = 0, 1$ but negative at $t = 1 / 2$ . Its degree-two Bernstein controls are $1 / 10, - 2 / 5, 1 / 10$ ; the negative middle control explains why the all-controls-nonnegative hypothesis in Bernstein control values bound a piecewise polynomial does not apply.

Facts & Assumptions

Given: The interval $[0, 1]$ , the polynomial $p (t) = t^{2} - t + 1 / 10$ , and exact real arithmetic.

[F1]

Piecewise Bernstein representation: a degree-two Bernstein polynomial has the form $c_{0} (1 - t)^{2} + 2 c_{1} t (1 - t) + c_{2} t^{2}$ .

[F2]

Bernstein control values bound a piecewise polynomial: nonnegative Bernstein controls suffice for nonnegativity throughout the interval.

[F3]

The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ : $(\binom{n}{1}) = n$ for every natural $n$ .

[F4]

The canonical natural $ι (n) = n \cdot 1_{F}$ of a field: the canonical embedding sends zero to zero and each successor to the previous image plus one.

Verification

1.1F1F3F4givenalgebra

By [F3], $(\binom{2}{1}) = 2$ ; applying the recursion in [F4] twice shows its canonical real value is $2$ . Expanding $\frac{1}{10} (1 - t)^{2} - \frac{4}{5} t (1 - t) + \frac{1}{10} t^{2}$ gives $t^{2} - t + 1 / 10$ , so [F1] identifies its Bernstein controls as $c_{0} = 1 / 10$ , $c_{1} = - 2 / 5$ , $c_{2} = 1 / 10$ .

1.2givenalgebra

Direct substitution gives $p (0) = p (1) = 1 / 10 > 0$ , while $p (1 / 2) = 1 / 4 - 1 / 2 + 1 / 10 = - 3 / 20 < 0$ . Thus positive samples at the two endpoints do not control the interpolant between them.

2.1F2step 1.1step 1.2∎

The middle Bernstein control is negative, so the sufficient hypothesis in [F2] is absent; the negative value in step 1.2 is consistent with the control-hull theorem.

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