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.

From psi to the logarithmic integral

Example

Assume countable choice.

The transfer from a classical psi error to pi retains $π (x) - Li (x) = \frac{2}{\log 2} + \frac{E (x)}{\log x} + \int_{2}^{x} \frac{E (t)}{t \log^{2} t} d t, E (t) = θ (t) - t .$ Both the prime-power error and this error integral are absorbed into a decreased classical exponential rate.

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]

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 $x$ , $ψ (x) - θ (x) \leq K_{2} \sqrt{x}$ .

[F3]

Abel summation recovers the prime-counting function from theta: For every real $x \geq 2$ , $π (x) = θ (x) / \log x + \int_{2}^{x} θ (t) / (t \log^{2} t) d t$ .

[F4]

Logarithmic integral: For real $x \geq 2$ , $Li (x) = \int_{2}^{x} d t / \log t$ , with $Li (2) = 0$ .

Verification

1.1F1F2

Under countable choice, the supplied bounds $ψ (t) - t = O (t e^{- c \sqrt{\log t}})$ and $0 \leq ψ (t) - θ (t) = O (\sqrt{t} \log t)$ give $E (t) = O (t e^{- c^{'} \sqrt{\log t}})$ after decreasing the positive constant. The ratio of the prime-power error to $t e^{- c^{'} \sqrt{\log t}}$ is $(\log t) e^{- \log t / 2 + c^{'} \sqrt{\log t}}$ , which is bounded.

2.1F3F4step 1.1algebra∎

Substitute $θ (t) = t + E (t)$ into the partial-summation identity. The main term $x / \log x + \int_{2}^{x} d t / \log^{2} t$ equals $Li (x) + 2 / \log 2$ by integration by parts. In the E-integral, [2,square root x] contributes O(square root x), and [square root x,x] contributes $O (x e^{- (c^{'} / \sqrt{2}) \sqrt{\log x}})$ . The endpoint term satisfies the same bound. At x=2 the empty integral leaves $2 / \log 2 + (\log 2 - 2) / \log 2 = 1 = π (2)$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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)