Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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  • 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.
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Re(1/z) is harmonic on a punctured disc and does not extend harmonically across 0

Statement refuted

Refuted claim: every harmonic function on a punctured disc extends harmonically across the puncture.

The witness is

u(z)=Re(1/z).

It is harmonic on 0<z<1, but it is unbounded near 0 and therefore does not extend harmonically there.

Facts & Assumptions

Given: The function f(z)=1/z on 0<z<1 and its real part u(z)=Re(f(z)).

[L3]

A bounded harmonic function near an isolated puncture does extend harmonically (A bounded harmonic function near an isolated puncture extends harmonically).

Counterexample

technique · direct
1.1

By [L1], the function 1/z is holomorphic on 0<z<1, so [L2] makes u(z)=Re(1/z) harmonic there.

L1L2
1.2

On the positive real axis, u(t)=1/t+ as t0, so u is unbounded near 0. If u had a harmonic extension across 0, it would be bounded on some small closed disc around 0, contradicting [L3].

L3algebra
2.1

Therefore u is harmonic on the punctured disc but does not extend harmonically across the puncture.

step 1.1step 1.2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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