Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07
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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.
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An inward field without a negative-time flow in the half-line

Statement refuted

On [0,), X=x is inward at 0 but has no negative-time flow through 0 staying in the half-line.

Facts & Assumptions

Given: The half-line M=[0,) with boundary point 0 and the constant smooth vector field X=x.

[L1]

In a boundary chart with half-space coordinate increasing into the manifold, a vector is inward exactly when its last coordinate is positive (Inward, outward, and boundary-tangent vectors).

Counterexample

technique · direct
1.1

In the identity boundary chart on M, the vector X0=x has coordinate +1, so it is inward at 0 by [L1].

givenL1algebra
2.1

Any integral curve through 0 must satisfy γ(t)=1 and γ(0)=0, hence γ(t)=t. It lies in M for t0 but not for any t<0. Therefore no flow through 0 can be defined for negative time while remaining in the half-line.

givenstep 1.1algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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