Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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.

Green's theorem is the curl statement for a planar field lifted to R3

Statement

Let D be a finite elementary Green region with its supplied decomposition, positively oriented, and let P,Q be C1 on an open UR2 containing D. Define the lift

F~(x,y,z):=(P(x,y), Q(x,y), 0)((x,y,z)U×R),

a field on the open set U×RR3. Then F~ is C1, its curl has first and second coordinates identically 0 and third coordinate xQyP at every point, independent of z, and the circulation of the planar field around the positive boundary chain equals the integral of the third coordinate of the curl of the lift:

D(P,Q)dr=D(curlF~)z(x,y,0)dA.

Facts & Assumptions

Given: The finite elementary Green region D with its supplied decomposition and positive orientation, the C1 functions P,Q on the open UD, and the lift F~ of the Statement.

[F1]

The curl of a C1 field on an open subset of R3 is curlG=(yGzzGy, zGxxGz, xGyyGx) (Divergence and curl of a C1 vector field).

[F2]

A map is of class Ck when each component is, a scalar component being C1 when its first partial derivatives exist and are continuous (Ck Euclidean maps and diffeomorphisms).

[F3]

For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, and DGdr and DPdx+Qdy denote that sum for the field (P,Q) (Positive orientation of elementary-region boundaries, Scalar line integrals with respect to arc length and vector-field line integrals).

[F4]

A finite elementary Green region is a nonempty finite union of elementary Green regions with pairwise disjoint interiors and the stated shared-arc conditions, supplied as data (Type I, Type II, and elementary regions for Green's theorem).

[L1]

Let D=D1DN be a finite elementary Green region with its supplied decomposition, oriented positively, and let P,Q be C1 on an open neighbourhood of D. Then DPdx+Qdy=D(xQyP)dA (Green's theorem for finite unions of elementary regions).

Proof

technique · direct
1.1

The three components of F~ are (x,y,z)P(x,y), (x,y,z)Q(x,y) and the constant 0. Their first partial derivatives are xF~x=xP, yF~x=yP, zF~x=0; xF~y=xQ, yF~y=yQ, zF~y=0; and all three of xF~z, yF~z, zF~z are 0. Each of these exists and is continuous on U×R because P and Q are C1 on U, so F~ is C1 there by [F2].

givenF2F5
2.1

By [F1] and step 1.1 the three coordinates of curlF~ are yF~zzF~y=00=0, then zF~xxF~z=00=0, and then xF~yyF~x=xQyP. All three are computed, and the third depends only on (x,y), so its value at (x,y,z) is its value at (x,y,0).

step 1.1F1
3.1

By [F4] the region D carries its supplied decomposition and P,Q are C1 on the open neighbourhood U of D, so [L1] gives DPdx+Qdy=D(xQyP)dA; by [F3] the left side is D(P,Q)dr, and by step 2.1 the integrand on the right is (curlF~)z(x,y,0). That is the asserted identity, and step 2.1 is the assertion about the three curl coordinates.

step 2.1F3F4L1

Remarks

  • This is a dictionary, not a new theorem. Both sides are the two sides of Green's theorem, rewritten. What the corollary records is that the planar integrand xQyP is a curl, so that the planar and the spatial developments on this page speak about one operator rather than two unrelated ones.

  • The route is deliberately one-way. The classical Stokes theorem for a C2 patch over a finite elementary Green region is proved from Green's theorem, so re-deriving Green's theorem from it would be circular. Nothing above uses Stokes' theorem.

Depends on

Used by

Dependency tree · two levels

47 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