Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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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.

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The weak inequality |f-g| <= |g| does not suffice in Rouche

Statement refuted

Refuted claim: Rouché's theorem remains valid when the strict boundary inequality ∣f−g∣<∣g∣ is weakened to ∣f−g∣≤∣g∣.

Facts & Assumptions

Given: The unit circle, g(z)=z, and f(z)=z+1.

[L1]

The classical theorem requires the strict inequality (Rouche's theorem in the classical strict-inequality form).

Counterexample

technique · direct
1.1

On ∣z∣=1 one has ∣f(z)−g(z)∣=∣1∣=1=∣z∣=∣g(z)∣, so the weak inequality holds everywhere on the boundary.

givenalgebra
2.1

But f vanishes at z=−1, which lies on the boundary itself. So the interior zero-count conclusion is no longer even well posed. This is exactly why [L1] is stated with a strict inequality.

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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