Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06‡ 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 Grothendieck ring of a tensor category is always commutative

Statement

False claim. The Grothendieck ring of a tensor category is always commutative.

Facts & Assumptions

Given: A tensor category.

[F1]

Duality gives an anti-isomorphism, reversing product order (Duality induces an anti-isomorphism on the Grothendieck ring).

Refutation

technique · direct
1.1

Let G=S3 and let VecG be the category of finite-dimensional G-graded vector spaces. Its simple objects δg are indexed by g∈G, and δg⊗δh≅δgh; it is a tensor category.

givenconstruct
2.1

Thus K0(VecG)≅Z[G]. Since S3 is nonabelian, for example (12)(23)≠(23)(12), this ring is not commutative. The order reversal in [F1] is consistent with, but does not remove, this counterexample.

step 1.1F1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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