ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Stokes on an interval with both endpoint chart signs

Example

For $f (t) = t^{2}$ on the increasingly oriented interval $[0, 1]$ , $\int_{[0, 1]} d f = 1 = f (1) - f (0) .$ The right endpoint chart $u = 1 - t$ is negative; its chart sign must be retained in the upper-half-line calculation.

Facts & Assumptions

[F1]

Stokes agrees with the fundamental theorem of calculus: For $a < b$ , orient $[a, b]$ increasingly. Every smooth $f$ on this interval satisfies $\int_{[a, b]} d f = f (b) - f (a),$ where the boundary point signs are $- 1$ at $a$ and $+ 1$ at $b$ . This agrees with the Riemann fundamental theorem of calculus.

[F2]

Compact-support Stokes on the upper half-space: Give $H^{n} = {x_{n} \geq 0}$ the standard orientation, $n \geq 1$ , and its face the outward-normal-first orientation. If $η \in Ω_{c}^{n - 1} (H^{n})$ and $j : \partial H^{n} ↪ H^{n}$ , then $\int_{H^{n}} d η = \int_{\partial H^{n}} j^{*} η .$ With $η = \sum_{i} a_{i} d x^{1} \land \dots \land \hat{d x^{i}} \land \dots \land d x^{n}$ , both sides are $(- 1)^{n} \int_{R^{n - 1}} a_{n} (x^{'}, 0) d x^{'}$ for $n > 1$ , and $- a_{1} (0)$ for $n = 1$ .

Verification

Given: The objects and hypotheses in the statement above.

1.1

The interval formula gives $\int d f = \int_{0}^{1} 2 t d t = 1$ and boundary values $(- 1) f (0) + (+ 1) f (1) = 1$ . These are induced endpoint signs, not unsigned point counting.

2.1

At the left endpoint $u = t$ is positive and the half-line boundary sign is negative. At the right endpoint $u = 1 - t$ is negative: the half-line calculation contributes $- f (1)$ , and the chart sign $- 1$ changes it to $+ f (1)$ . More explicitly apply that local calculation to partition-weighted f supported near the endpoint; the two signs multiply in exactly this way. Thus the local calculation reproduces both endpoint values, including the zero value at t=0.

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Lee Example 16.12 and p.405 negative-chart explanation (standard reference, not scraped)