Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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.

One dimension shift along an injective copresentation

Example

Assume the Axiom of Dependent Choice.

Let I be supplied injective resolution data on a class D of objects of A, let F:A→B be additive and left exact, and let 0→A→J→C→0 be a short exact sequence in D with J injective. Then for every n>1 the connecting map gives an isomorphism RIn−1F(C)→ ∼ RInF(A).

Facts & Assumptions

Given: A short exact sequence 0→A→J→C→0 in D with J injective and an integer n>1.

[L1]

The right derived functors form a cohomological delta functor (Right derived functors form a cohomological delta functor).

[L2]

Positive right derived functors vanish on injective objects (Positive right derived functors vanish on injective objects).

[L3]

Vanishing of the adjacent injective terms makes the connecting map an isomorphism (Dimension shift for a cohomological delta functor effaced in the middle).

Verification

technique · direct
1.1L1givenconstruct

By [L1], the given short exact sequence yields an exact segment RIn−1F(J)→RIn−1F(C)→∂n−1RInF(A)→RInF(J).

2.1L2L3step 1.1∎

Since J∈D is injective and n>1, [L2] gives RIn−1F(J)=RInF(J)=0. Hence [L3] makes the connecting map in step 1.1 an isomorphism.

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