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.

Optimizing the prime number theorem contour height

Example

Assume countable choice.

At $T = \exp (A \sqrt{\log x})$ for fixed $A > 0$ , the exponential rates of the finite-zero and truncation terms in the explicit-formula error balance are respectively $c_{0} / A$ and $A$ . They therefore have the same square-root-logarithm scale, optimized when $A = \sqrt{c_{0}}$ .

Facts & Assumptions

Given: Countable choice and the data and hypotheses of the statement.

[F1]

Chebyshev psi prime number theorem error: There is an absolute $c > 0$ such that for $x \geq 2$ , $ψ (x) = x + O (x e^{- c \sqrt{\log x}}) .$

[F2]

Zeta explicit formula zero free error balance: For $x \geq 2$ and finite $T \geq 3$ , the classical region and truncated explicit formula give $ψ (x) - x = O (x e^{- c_{0} \log x / \log (T + 2)} \log^{2} T + \frac{x \log^{2} (x T)}{T} + \log x) .$ Constants may be enlarged and the positive region constant decreased. The zero sum used in the proof is finite.

Verification

1.1F2algebra

Under countable choice, put $u = \sqrt{\log x}$ . The error-balance lemma gives respectively $O (x u^{2} e^{- (c_{0} / A) u + o (1)})$ , $O (x u^{4} e^{- A u})$ , and $O (u^{2})$ . Thus any $c < \min (c_{0} / A, A)$ absorbs all polynomial factors. Equality of the two exponential rates occurs at $A = \sqrt{c_{0}}$ ; a fixed positive $A$ already suffices.

2.1F2F1step 1.1algebra∎

Thus the choice $A = \sqrt{c_{0}}$ balances the two exponential rates at $\sqrt{c_{0}}$ . More generally any fixed $A > 0$ gives a positive decay rate $\min (c_{0} / A, A)$ ; bounded x can use $T = 3$ and an enlarged constant. This is the square-root-logarithm decay scale recorded in the theorem-level estimate [F1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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 6.9 proof (standard reference, not scraped)