ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-09-24 (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.

Zero free region parameter balance

Example

Let $ρ = β + i γ$ be a zero of $ζ$ with $β < 1$ and $∣ γ ∣ \geq 3$ . In the high-height zero-free-region proof, suppose $\frac{4}{1 + δ - β} \leq \frac{3}{δ} + C \log (∣ γ ∣ + 2),$ holds at $δ = 1 / (2 C \log (∣ γ ∣ + 2))$ , where $C$ is a fixed sufficiently large positive comparison constant. Then $1 - β \geq 1 / (14 C \log (∣ γ ∣ + 2))$ .

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Riemann zeta classical zero free region: Assuming countable choice, there is an absolute $c_{0} > 0$ such that $ζ$ has no zeros in $σ \geq 1 - c_{0} / \log (∣ t ∣ + 2)$ . The pole at $s = 1$ is not a zero.

Verification

1.1givenalgebra

Write $L = \log (∣ γ ∣ + 2)$ . Here $L > 0$ and $C > 0$ , so the chosen $δ$ is positive. Apply the displayed assumed inequality at this value of $δ$ . Substitution makes $3 / δ + C L = 7 C L$ , so $1 + δ - β \geq 4 / (7 C L)$ ; the denominator is positive because $β < 1$ .

2.1F1step 1.1algebra∎

Subtract $δ = 1 / (2 C L)$ to obtain $(4 / 7 - 1 / 2) / (C L) = 1 / (14 C L)$ . A smaller constant proves exclusion on a closed boundary. This calculation applies only to $∣ γ ∣ \geq 3$ ; under countable choice, [F1] states the all-height zero-free region, whose small-height conclusion is not supplied by this conditional calculation.

Depends on

Riemann zeta classical zero free region

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theorem 6.6 proof (standard reference, not scraped)