ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

A directed union of subgroups distributes over intersection with a fixed subgroup

Example

Let $A = ⨁_{n \geq 1} Z e_{n}$ , let $B_{n} = ⟨ e_{1}, \dots, e_{n} ⟩$ , and let $C = ⟨ e_{1} + e_{2}, e_{3}, e_{4}, \dots ⟩$ . Then $(B_{n})$ is directed, $⋁_{n} B_{n} = A$ , and

$(⋁_{n} B_{n}) \land C = C = ⋁_{n} (B_{n} \land C) .$

This is the AB5 lattice identity in a concrete module calculation.

Facts & Assumptions

Given: The subgroup chain $(B_{n})$ and the fixed subgroup $C$ displayed in the statement.

[L1]

AB5 is the directed-join distributivity law for subobjects (The axioms AB5 and AB5*).

[L2]

Module categories are Grothendieck, hence satisfy AB5 (Module categories are Grothendieck categories).

Verification

technique · direct

1.1

The subgroups $B_{n}$ are directed by inclusion and their union is all of $A$ , since every element of $A$ has finite support. Also $B_{n} \cap C = ⟨ e_{1} + e_{2}, e_{3}, \dots, e_{n} ⟩$ for $n \geq 2$ , because those are exactly the generators of $C$ lying in the first $n$ coordinates.

L1algebra

2.1

Taking the union of the intersections from step 1.1 recovers all of $C$ , so $⋁_{n} (B_{n} \land C) = C = (⋁_{n} B_{n}) \land C .$ This is precisely the AB5 identity [L1], exactly as the abstract theorem [L2] predicts for module categories.

L1L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Alexandre Grothendieck, Sur quelques points d'algèbre homologique, Barr translation, Section 1.5 (standard reference, not scraped)