Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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.

A fraction r/s is a unit in S−1R exactly when ar∈S for some a∈R

Statement

Let R be a commutative ring and S⊆R multiplicative. A fraction r/s∈S−1R is a unit if and only if ar∈S for some a∈R.

Facts & Assumptions

Given: A fraction r/s in S−1R.

[F1]

Fractions satisfy the usual multiplication law, and equality is detected by an annihilating element of S (The localisation relation is an equivalence relation and fraction arithmetic is well defined, Equality, vanishing, and the kernel of the localisation map).

Proof

technique · direct
1.1

If ar∈S, then as/(ar) is a valid fraction and (r/s)(as/(ar))=ars/(sar)=1 by [F1]. Hence r/s is a unit.

F1
2.1

Conversely, suppose (r/s)(b/t)=1. Then rb/(st)=1, so [F1] supplies u∈S with u(rb−st)=0. Therefore (ub)r=ust∈S, because u,s,t∈S. Taking a=ub proves the criterion.

F1∎

Depends on

Used by

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