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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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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.

Divergence and curl are linear and satisfy the scalar product rules

Statement

Let n1, let URn be open, let F,G:URn be C1, let f:UR be C1 and let a,bR. Then aF+bG and fF are C1 on U and

div(aF+bG)=adivF+bdivG,

div(fF)=f,F+fdivF.

If moreover n=3, then

curl(aF+bG)=acurlF+bcurlG,curl(fF)=f×F+fcurlF.

Here div, curl and the coordinate naming are those of Divergence and curl of a C1 vector field, f is the gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, , is the inner product of The Euclidean inner product x,y=k<nxkyk on Rn and × is the cross product of The cross product in R3.

Facts & Assumptions

Given: The open set U, the C1 maps F,G:URn, the C1 scalar f:UR and the reals a,b of the Statement.

[F1]

The divergence of a C1 field F on an open URn is divF=i<niFi, and for n=3 its curl is curlF=(yFzzFy, zFxxFz, xFyyFx) (Divergence and curl of a C1 vector field).

[F2]

For scalar-valued f on an open subset of Rm, the gradient is f=(0f,,m1f) (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F3]

For u=(ux,uy,uz) and v=(vx,vy,vz) in R3, u×v=(uyvzuzvy,uzvxuxvz,uxvyuyvx) (The cross product in R3).

[F4]

For x,yRm, x,y=k<mxkyk (The Euclidean inner product x,y=k<nxkyk on Rn).

[F5]

A map f:URq is of class Ck when each component is of class Ck (Ck Euclidean maps and diffeomorphisms).

[L1]

For real functions of one real variable differentiable at a point, f+g is differentiable there with (f+g)=f+g, αf is differentiable there with (αf)=αf, and fg is differentiable there with (fg)=fg+fg (Sums, scalar multiples, products and quotients: (f+g)(c)=f(c)+g(c), (αf)(c)=αf(c), (fg)(c)=f(c)g(c)+f(c)g(c), and (f/g)(c)=(f(c)g(c)f(c)g(c))/g(c)2 when g(c)0).

Proof

technique · direct
1.1

A partial derivative j at a point p is the ordinary one-variable derivative at 0 of t()(p+tej), so [L1] applies to it verbatim: for scalar C1 functions u,v on U and reals a,b one has j(au+bv)=aju+bjv and j(uv)=(ju)v+ujv pointwise on U, and the right-hand sides are continuous, so au+bv and uv are again C1.

givenL1F5
1.2

Applying 1.1 componentwise, aF+bG and fF have C1 components, hence are C1 by [F5]; so all four expressions in the Statement are defined.

givenL1F5
2.1

By [F1], div(aF+bG)=i<ni(aFi+bGi), and step 1.1 rewrites each summand as aiFi+biGi; summing gives ai<niFi+bi<niGi=adivF+bdivG.

step 1.1F1algebra
2.2

By [F1], the first coordinate of curl(aF+bG) is y(aFz+bGz)z(aFy+bGy), which step 1.1 rewrites as a(yFzzFy)+b(yGzzGy); the second coordinate is z(aFx+bGx)x(aFz+bGz)=a(zFxxFz)+b(zGxxGz) and the third is x(aFy+bGy)y(aFx+bGx)=a(xFyyFx)+b(xGyyGx). The three coordinates are those of acurlF+bcurlG.

step 1.1F1algebra
2.3

By [F1], div(fF)=i<ni(fFi), and step 1.1 rewrites each summand as (if)Fi+fiFi. Splitting the sum gives i<n(if)Fi+fi<niFi, whose first term is f,F by [F2] and [F4] and whose second is fdivF by [F1].

step 1.1F1F2F4algebra
2.4

By [F1], the first coordinate of curl(fF) is y(fFz)z(fFy), which step 1.1 rewrites as ((yf)Fz(zf)Fy)+f(yFzzFy). By [F2] and [F3] with u=f and v=F, the first bracket is the first coordinate uyvzuzvy of f×F, and the second summand is f times the first coordinate of curlF.

step 1.1F1F2F3algebra
2.5

By [F1], the second coordinate of curl(fF) is z(fFx)x(fFz), which step 1.1 rewrites as ((zf)Fx(xf)Fz)+f(zFxxFz). By [F2] and [F3] the first bracket is the second coordinate uzvxuxvz of f×F, and the second summand is f times the second coordinate of curlF.

step 1.1F1F2F3algebra
2.6

By [F1], the third coordinate of curl(fF) is x(fFy)y(fFx), which step 1.1 rewrites as ((xf)Fy(yf)Fx)+f(xFyyFx). By [F2] and [F3] the first bracket is the third coordinate uxvyuyvx of f×F, and the second summand is f times the third coordinate of curlF.

step 1.1F1F2F3algebra
3.1

Steps 2.4, 2.5 and 2.6 give the three coordinates of f×F+fcurlF, so curl(fF)=f×F+fcurlF; with steps 2.1, 2.2 and 2.3 this is every assertion of the Statement.

step 2.1step 2.2step 2.3step 2.4step 2.5step 2.6

Remarks

Depends on

Used by

Dependency tree · two levels

33 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