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.

The volume contradiction for an alleged Banach–Tarski decomposition

Example

Write the finite-additivity calculation for a positive-radius ball $K$ .

Facts & Assumptions

Given: $K = ⨆_{i < m} A_{i}$ and two disjoint copies $K_{0}, K_{1}$ reassembled as $K_{0} ⊔ K_{1} = ⨆_{i < m} g_{i} A_{i}$ .

[F1]

The Solovay model has no Banach–Tarski decomposition: states that the alleged reassembly cannot exist.

[F2]

Universal real measurability transfers to finite-dimensional Euclidean spaces: every piece $A_{i} \subseteq R^{3}$ is measurable.

[F3]

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation and Lebesgue measure on $R^{n}$ is invariant under every orthogonal linear map: rigid motions preserve measure.

[F4]

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included: inner and outer cubes give $0 < λ (K) < \infty$ for positive radius.

[F5]

The Solovay inner model satisfies Dependent Choice and AC implies DC implies countable choice: $M$ satisfies DC and hence Countable Choice, the hypothesis required by the orthogonal-invariance and box-measure results.

Verification

1.1

F5 supplies Countable Choice inside $M$ . By F2 and F4 all terms are measurable and $0 < V = λ (K) < \infty$ . Finite additivity gives $V = \sum_{i < m} λ (A_{i})$ . F3 gives $\sum_{i < m} λ (g_{i} A_{i}) = \sum_{i < m} λ (A_{i}) = V$ .

F2F3F4F5

2.1

But disjoint congruent copies give $λ (K_{0} ⊔ K_{1}) = λ (K_{0}) + λ (K_{1}) = V + V = 2 V$ . The same target set was the reassembly in step 1.1, so $V = 2 V$ , hence $V = 0$ , contradicting $0 < V < \infty$ and verifying F1's exclusion by the promised calculation.

F1F3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

45 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.