Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30 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.

The snake lemma applied to multiplication by an integer

Example

Fix n1 and apply multiplication by n to the short exact sequence 0Z×nZZ/n0. The snake lemma produces 000Z/nδZ/n0Z/n1Z/n0, so the connecting morphism is an isomorphism, and under the standard identifications it is the identity.

Facts & Assumptions

Given: The multiplication-by-n endomorphism of the short exact sequence above.

[L1]

Abelian groups form an abelian category (Abelian groups form an abelian category).

[L2]

The snake lemma gives the exact sequence attached to that ladder (Snake lemma in an abelian category).

Verification

1.1

Multiplication by n on Z has zero kernel and cokernel Z/n, while the induced map on the quotient term Z/n is zero. The induced map coker(×n)coker(×n) is multiplication by n on Z/n, hence is 0, and the induced map coker(×n)coker(0) is the identity of Z/n.

L1algebra
2.1

Substituting those terms into [L2] gives the displayed exact sequence. Exactness at the first copy of Z/n forces δ to be an isomorphism. In the standard snake construction, the class of 1 in ker(0)=Z/n lifts to 1Z and then maps to the class of 1 in coker(×n)=Z/n, so under these standard identifications δ is the identity.

L2step 1.1algebra
3.1

This is the concrete snake sequence for multiplication by an integer.

step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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