Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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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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False: vector line integrals are invariant under reversing a path

Statement

For every continuous vector field F and piecewise-C1 path γ on whose trace it is defined, γFdr=γFdr.

Facts & Assumptions

Given: The constant field F=(1,0) and the segment γ(t)=(t,0) on [0,1].

[L1]

On a C1 path η, the vector line integral is the integral of F(η(t)),η(t) (Scalar line integrals with respect to arc length and vector-field line integrals).

[L2]

An orientation-reversing reparametrization negates a vector line integral but leaves a scalar ds integral unchanged (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

Refutation

technique · direct
1.1

Since γ(t)=(1,0), [L1] and [L3] give γFdr=011dt=1.

givenL1L3algebra
2.1

The reversal is orientation-reversing, so [L2] gives γFdr=1.

givenstep 1.1L2algebra
3.1

Since 11, the statement is false. The scalar analogue is true because [L2] says that ds removes the orientation sign.

step 1.1step 2.1L2

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: 84 results over 18 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