ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Initial terms of the square-root-two rotation

Example

Assume the Axiom of Countable Choice. For $α = \sqrt{2}$ , the fractional parts ${n α}$ for $n = 0, \dots, 5$ are approximately

$0, 0.4142, 0.8284, 0.2426, 0.6569, 0.0711.$

Weyl's theorem says that the full sequence is equidistributed modulo one.

Facts & Assumptions

Given: Countable choice and the positive square root $\sqrt{2}$ .

[F1]

The element $\sqrt{2}$ exists and is irrational ( $\sqrt{2}$ exists in every complete ordered field, and is irrational).

[F2]

Every irrational rotation sequence is equidistributed modulo one (Weyl equidistribution for irrational rotations), in the half-open interval sense of Equidistribution modulo one.

[F3]

Countable choice is the standing assumption required by [F2] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

technique · direct arithmetic followed by the theorem

1.1

The inequalities obtained by squaring positive rational bounds show $1 < \sqrt{2} < 3 / 2$ , $4 / 3 < \sqrt{2} < 5 / 3$ , $5 / 4 < \sqrt{2} < 3 / 2$ , and $7 / 5 < \sqrt{2} < 8 / 5$ . Hence the relevant integer parts of $n \sqrt{2}$ for $n = 0, \dots, 5$ are $0, 1, 2, 4, 5, 7$ .

F1algebra

2.1

Therefore the exact fractional parts are $0, \sqrt{2} - 1, 2 \sqrt{2} - 2, 3 \sqrt{2} - 4, 4 \sqrt{2} - 5, 5 \sqrt{2} - 7.$ Direct integer squaring gives ${(\frac{14142135}{10^{7}})}^{2} < 2 < {(\frac{14142136}{10^{7}})}^{2},$ so positivity places $\sqrt{2}$ between these two rational numbers; substituting the bounds into the five exact expressions and rounding to four decimal places gives the displayed list.

F1step 1.1algebra

3.1

Since $\sqrt{2}$ is irrational by [F1], [F2] applies and proves that the entire infinite sequence is equidistributed. The six computations in step 2.1 illustrate this orbit; they do not by themselves prove its asymptotic distribution. Countable choice is used only through [F2].

F1F2F3step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)