Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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 keyhole contour sees the two boundary values of z^(alpha-1)

Statement

Fix α∈C, and on the slit plane C∖[0,∞) define

zα−1:=exp⁡((α−1)Log⁡z)

with Arg⁡z∈(0,2π). Then for every x>0 the two boundary values on the positive axis satisfy

lim⁡y↓0(x+iy)α−1=xα−1,lim⁡y↓0(x−iy)α−1=e2πiαxα−1.

Facts & Assumptions

Given: A complex exponent α and the branch zα−1=exp⁡((α−1)Log⁡z) with Arg⁡z∈(0,2π).

Proof

technique · direct
1.1

On the upper lip of the slit one has Arg⁡(x+i0)=0, so Log⁡(x+i0)=log⁡x and therefore lim⁡y↓0(x+iy)α−1=exp⁡((α−1)log⁡x)=xα−1.

givenalgebra
2.1

On the lower lip one has Arg⁡(x−i0)=2π, so Log⁡(x−i0)=log⁡x+2πi. Hence lim⁡y↓0(x−iy)α−1=exp⁡((α−1)(log⁡x+2πi))=e2πi(α−1)xα−1=e2πiαxα−1.

step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

23 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