ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 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.

A polynomial potential evaluates work along every path by endpoints

Example

Let

$ϕ (x, y, z) = x^{2} y + y z^{2}, F = \nabla ϕ = (2 x y, x^{2} + z^{2}, 2 y z) .$

Every piecewise- $C^{1}$ path from $p = (0, 0, 0)$ to $q = (1, 2, 1)$ satisfies

$\int_{γ} F \cdot d r = 4.$

Facts & Assumptions

Given: The polynomial potential, field, and endpoints in the Example.

[L1]

The gradient theorem gives $\int_{γ} \nabla ϕ \cdot d r = ϕ (q) - ϕ (p)$ for every piecewise- $C^{1}$ path from $p$ to $q$ (The gradient theorem: the line integral of a gradient is the endpoint increment).

[L2]

On a $C^{1}$ path $γ$ , the vector line integral is the integral of $⟨ F (γ (t)), γ^{'} (t) ⟩$ (Scalar line integrals with respect to arc length and vector-field line integrals).

[L3]

The power rule gives $(t^{3})^{'} = 3 t^{2}$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term), and for a continuous function whose interior derivative admits an integrable extension, Newton-Leibniz integrates that extension to the endpoint increment (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

Verification

technique · direct

1.1

Coordinate differentiation gives the displayed gradient, and direct evaluation gives $ϕ (q) = 4$ and $ϕ (p) = 0$ .

givenalgebra

1.2

For the affine segment $γ (t) = (t, 2 t, t)$ , one has $F (γ (t)) = (4 t^{2}, 2 t^{2}, 4 t^{2})$ and $q - p = (1, 2, 1)$ . Thus [L2] gives the integrand $12 t^{2}$ .

givenL2algebra

2.1

Apply [L1] and step 1.1 to any path from $p$ to $q$ . Its integral is $4 - 0 = 4$ .

step 1.1L1algebra

3.1

Since $12 t^{2} = (4 t^{3})^{'}$ , [L3] gives $\int_{0}^{1} 12 t^{2} d t = 4$ , agreeing with step 2.1.

step 1.2L3algebra∎

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: 93 results over 21 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

J. Lebl, Basic Analysis II, section 9.3 (standard reference, not scraped)