Alphabeta Math
ExampleConstruction: AI-adaptedVerification: 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.

A two-term complex has kernel and cokernel homology

Example

Let R be a ring and let f:MN be a homomorphism. Regard 0MfN0 as a chain complex with M in degree 1 and N in degree 0. Then H1ker(f),H0coker(f), and all other homology objects are zero.

Facts & Assumptions

Given: A module homomorphism f:MN.

[L1]

Module categories are abelian (Modules over a ring form an abelian category).

[L2]

Cycles, boundaries, and homology are defined by kernels, images, and the quotient Zn/Bn (Cycle and boundary subobjects of a complex, Homology object of a chain complex).

Verification

technique · direct
1.1

In degree 1, the outgoing differential is f, so Z1=ker(f) and B1=0. In degree 0, the incoming differential is f and the outgoing one is 0, so Z0=N and B0=im(f). Every other term is zero.

L1L2givenalgebra
2.1

Therefore [L2] gives H1=ker(f),H0=N/im(f)=coker(f), and Hn=0 for n0,1.

L2step 1.1

Depends on

Used by

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