Alphabeta Math
CorollaryStatement: AI-adaptedProof: 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.

Prime number theorem

Statement

As x→∞, π(x)∼x/log⁡x,θ(x)∼x,ψ(x)∼x. These three asymptotic assertions are equivalent.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Prime number theorem arithmetic progressions: For each fixed q≥1 and each unit class a modulo q, π(x;q,a)∼Li⁡(x)/φ(q). In particular, the case q=1 gives π(x)∼Li⁡(x).

[F2]

Logarithmic integral asymptotic expansion: For each fixed integer m≥1, as x→∞, Li⁡(x)=∑j=0m−1j!xlog⁡j+1x+Om(xlog⁡m+1x).

[F3]

Abel summation recovers the prime-counting function from theta: For every real x≥2, π(x)=θ(x)log⁡x+∫2xθ(t)tlog⁡2t dt.

[F4]

Psi and theta differ by at most a square-root term: There are positive constants K1,K2 such that for every real x≥2, 0≤ψ(x)−θ(x)≤K1xlog⁡x and, for all sufficiently large x, ψ(x)−θ(x)≤K2x.

Proof

1.1F1F2

Apply the progression theorem at q=1, where every prime belongs to the unique unit class and φ(1)=1. Its counting asymptotic and the first Li term give π(x)∼x/log⁡x.

1.2F3

Independently, if θ(x)∼x, the exact Abel formula gives π(x)∼x/log⁡x: its integral is O(x/log⁡2x), by splitting at x and using θ(t)=O(t).

2.1F4step 1.1step 1.2∎

Conversely summing log⁡p=log⁡x−∫pxdt/t over the finitely many primes p≤x gives θ(x)=π(x)log⁡x−∫2xπ(t)dt/t. If π(t)∼t/log⁡t, the integral is O(x/log⁡x)=o(x) by the same square-root split, so θ(x)∼x. Finally 0≤ψ(x)−θ(x)=O(xlog⁡x)=o(x), proving both directions between theta and psi. Combine these implications with the first step.

Depends on

Used by

Dependency tree · two levels

17 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