ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01‡ 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.

An injective resolution of an abelian group beginning with a divisible group

Example

Assume the Axiom of Choice through Baer's criterion.

For the abelian group $Z$ , the sequence $0 \to Z \to Q \to Q / Z \to 0$ is an injective resolution. It begins with the standard embedding of $Z$ into the divisible group $Q$ .

Facts & Assumptions

Given: The abelian group $Z$ .

[L1]

Every module admits an injective resolution (Every module admits an injective resolution).

[L2]

Every abelian group embeds in a divisible abelian group (Every abelian group embeds in a divisible abelian group).

[L3]

Over $Z$ , injective modules are exactly divisible groups (Over a PID, injective modules are exactly divisible modules).

Verification

technique · direct

1.1

The inclusion $Z ↪ Q$ is injective, its cokernel is $Q / Z$ , and both $Q$ and $Q / Z$ are divisible. Therefore both are injective by [L3], and the displayed sequence is exact.

L3givenalgebra

2.1

Thus the sequence is already an injective resolution of $Z$ . It starts with an embedding into the divisible group promised by [L2], and it is a concrete instance of the general existence statement [L1].

L1L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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 A. Weibel, An Introduction to Homological Algebra (standard reference, not scraped)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)