ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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 closed form and its potential

Example

On $C^{2}$ , let $g = 2 {\overset{ˉ}{z}}_{1} d {\overset{ˉ}{z}}_{1} + z_{1} d {\overset{ˉ}{z}}_{2}, u = {\overset{ˉ}{z}}_{1}^{2} + z_{1} {\overset{ˉ}{z}}_{2} .$ Then $\overset{ˉ}{\partial} g = 0$ and $\overset{ˉ}{\partial} u = g$ .

Facts & Assumptions

Given: The displayed polynomial $(0, 1)$ -form $g$ and function $u$ on all of $C^{2}$ .

[F1]

The coordinate Wirtinger derivatives are $\partial_{z_{k}} = \frac{1}{2} (\partial_{x_{k}} - i \partial_{y_{k}})$ and $\partial_{{\overset{ˉ}{z}}_{k}} = \frac{1}{2} (\partial_{x_{k}} + i \partial_{y_{k}})$ (Wirtinger operators in $C^{m}$ ).

[F2]

The $\overset{ˉ}{\partial}$ coefficient formula differentiates each coefficient in ${\overset{ˉ}{z}}_{j}$ and wedges $d {\overset{ˉ}{z}}_{j}$ before the existing type factors (Bigraded complex forms and the Dolbeault operators).

Proof

technique · direct

1.1F1F2givenalgebra

By [F1], $\partial_{{\overset{ˉ}{z}}_{1}} u = 2 {\overset{ˉ}{z}}_{1}$ and $\partial_{{\overset{ˉ}{z}}_{2}} u = z_{1}$ : the cross derivatives $\partial_{{\overset{ˉ}{z}}_{1}} (z_{1} {\overset{ˉ}{z}}_{2})$ and $\partial_{{\overset{ˉ}{z}}_{2}} ({\overset{ˉ}{z}}_{1}^{2})$ are zero. The scalar case of [F2] therefore gives $\overset{ˉ}{\partial} u = 2 {\overset{ˉ}{z}}_{1} d {\overset{ˉ}{z}}_{1} + z_{1} d {\overset{ˉ}{z}}_{2} = g$ .

2.1F1F2givenalgebra∎

For $g = g_{1} d {\overset{ˉ}{z}}_{1} + g_{2} d {\overset{ˉ}{z}}_{2}$ , where $g_{1} = 2 {\overset{ˉ}{z}}_{1}$ and $g_{2} = z_{1}$ , [F1] gives $\partial_{{\overset{ˉ}{z}}_{1}} g_{1} = 2$ and $\partial_{{\overset{ˉ}{z}}_{2}} g_{1} = \partial_{{\overset{ˉ}{z}}_{1}} g_{2} = \partial_{{\overset{ˉ}{z}}_{2}} g_{2} = 0$ . Hence [F2] gives $\overset{ˉ}{\partial} g = 2 d {\overset{ˉ}{z}}_{1} \land d {\overset{ˉ}{z}}_{1} = 0$ by alternation; in particular, both mixed cross derivatives vanish.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Lebl, Tasty Bits of Several Complex Variables, v4.4, Chapter 4 §4.4 (standard reference, not scraped)