Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.

False: every closed C1 field on a connected open set is exact

Statement

Every closed C1 vector field on a connected open subset of Rn is exact.

Facts & Assumptions

Given: The proposed implication, the punctured plane U=R2{0}, and the vortex field F(x,y)=(yx2+y2,xx2+y2).

[L1]

With coordinates indexed from 0, so that F=(F0,F1), closedness means yF0=xF1, while exactness supplies a C2 potential with gradient F (Exact and closed C1 vector fields).

[L2]

A gradient line integral is its potential's endpoint increment and is therefore zero on a closed path (The gradient theorem: the line integral of a gradient is the endpoint increment).

[L3]

Vector line integrals integrate F(γ(t)),γ(t); sine and cosine have derivatives cost and sint and satisfy sin2t+cos2t=1 (Scalar line integrals with respect to arc length and vector-field line integrals, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L6]

On a star-shaped open domain, closedness and exactness are equivalent for C1 vector fields (On a star-shaped open domain, closed, exact, conservative, path-independent, and zero-loop are equivalent).

Refutation

technique · direct
1.1

Direct differentiation gives yF0=(y2x2)/(x2+y2)2=xF1, so [L1] makes F closed and C1 on the open set U. Radial segments at positive radius followed by a circular arc give a piecewise-C1 path in U between any two of its points.

givenL1algebra
1.2

For γ(t)=(cost,sint) on [0,2π], [L3] gives F(γ(t))=γ(t)=(sint,cost). Hence [L3] and [L4] give γFdr=02π1dt=2π0.

givenL3L4algebra
2.1

If U=AB were a separation into disjoint nonempty relatively open sets, choose uA and vB and let η:[0,1]U be the path from step 1.1. The function equal to 0 when η(t)A and to 1 when η(t)B is locally constant, hence continuous, and has endpoint values 0 and 1. By [L5] it would take the value 1/2, which is impossible. Thus U is connected.

step 1.1L5choosealgebra
3.1

If F were exact, [L1] and [L2] would make the integral in step 1.2 zero. Thus it is not exact, and the statement is false despite steps 1.1 and 2.1. The valid correction in [L6] replaces connectedness by the stronger star-shaped hypothesis.

step 1.1step 2.1step 1.2L1L2L6

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 167 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources