ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Chacon spacer measure budget

Example

Assume AC for the Lebesgue-measure interpretation. The initial Chacon column has mass $2 / 3$ . The first two spacers have masses $2 / 9$ and $2 / 27$ . All spacers together have mass $1 / 3$ ; after stage $r \geq 0$ the unused reservoir has mass $3^{- (r + 1)} \to 0$ . Spacers are retained, so it is the unused tail, not total spacer mass, that tends to zero.

Facts & Assumptions

[F1]

Spacer $J_{j}$ has width $2 / 3^{j + 2}$ , the initial column has width $2 / 3$ , and all the physical spacers are disjoint and retained Chacon three cut one spacer towers.

[F2]

Assume AC for the measure assertions in F1 The Axiom of Choice.

Verification

Given: The finite normalized Chacon construction.

1.1F1F2

At the first cut the spacer has width $2 / 3^{2} = 2 / 9$ , and at the second it has width $2 / 3^{3} = 2 / 27$ . After stage $r$ the retained spacer mass is the finite sum $S_{r} = \sum_{j = 0}^{r - 1} 2 / 3^{j + 2}$ . The empty sum at $r = 0$ is zero. For $r \geq 1$ , multiplying the sum by $1 - 1 / 3$ cancels all interior terms, giving $(2 / 3) S_{r} = (2 / 9) (1 - 3^{- r})$ , hence $S_{r} = 1 / 3 - 3^{- (r + 1)}$ . The formula also gives zero at $r = 0$ .

2.1F1F2step 1.1∎

Adding the initial mass yields $2 / 3 + S_{r} = 1 - 3^{- (r + 1)}$ , whose missing mass is exactly the reservoir. As $r \to \infty$ , $S_{r} \to 1 / 3$ and the reservoir tends to zero. Countable additivity on the disjoint spacer intervals identifies their union's measure with this sum; their union is also the physical interval $[2 / 3, 1)$ , since their adjacent endpoints tend to one. The numerical geometric identities themselves are choice-free; AC is used only through F1's Lebesgue-measure interpretation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Sarig Problem 3.8 pp.99–101 (standard reference, not scraped)