CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Chebyshev theta prime number theorem error

Statement

For some absolute $c > 0$ and all $x \geq 2$ , $θ (x) = x + O (x e^{- c \sqrt{\log x}}) .$

Facts & Assumptions

Given: 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]

Psi and theta differ by at most a square-root term: There are positive constants $K_{1}, K_{2}$ such that for every real $x \geq 2$ , $0 \leq ψ (x) - θ (x) \leq K_{1} \sqrt{x} \log x$ and, for all sufficiently large $x$ , $ψ (x) - θ (x) \leq K_{2} \sqrt{x} .$

Proof

1.1

The comparison gives $0 \leq ψ (x) - θ (x) \leq K \sqrt{x} \log x$ . Thus $∣ θ (x) - x ∣ \leq ∣ ψ (x) - x ∣ + K \sqrt{x} \log x$ .

2.1

The first term has the asserted bound. Writing $u = \sqrt{\log x}$ , the ratio of the second to $x e^{- c u}$ is $u^{2} e^{- u^{2} / 2 + c u}$ , bounded on $u \geq \sqrt{\log 2}$ . This proves the result after enlarging the constant.

F1step 1.1∎

Depends on

Used by

Prime number theorem logarithmic integral Theorem

Dependency tree · two levels

8 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, equation (6.13), pp.179–181 (standard reference, not scraped)