Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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.

z↦z2 is entire with derivative 2z, directly from the complex difference quotient

Example

The square function f(z)=z2 is entire and satisfies f′(z)=2z. In Cartesian coordinates its components are u(x,y)=x2−y2,v(x,y)=2xy, and both are harmonic; v is a harmonic conjugate of u.

Facts & Assumptions

Given: An arbitrary z∈C.

[L1]

A function is complex differentiable at z when its punctured-domain difference quotient has a complex limit there, and it is entire when it is complex differentiable at every point of C (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L2]

The complex modulus is multiplicative, satisfies the triangle inequality, and is definite (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[L3]

For a holomorphic function with C2 components, both components satisfy Laplace's equation, and the imaginary component is a harmonic conjugate of the real component (The C2 real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

Verification

technique · direct computation
1.1

For every nonzero increment h, f(z+h)−f(z)h=(z+h)2−z2h=2z+h.

algebra
2.1

Since ∣(2z+h)−2z∣=∣h∣→0, the quotient in step 1.1 tends to 2z. Thus f′(z)=2z by [L1].

step 1.1L1L2
3.1

The point z was arbitrary, so f is entire.

step 2.1L1
3.2

Writing z=x+iy gives z2=(x2−y2)+i(2xy). Hence ux=2x=vy and uy=−2y=−vx, in agreement with step 2.1.

step 2.1algebra
4.1

Moreover uxx+uyy=2−2=0 and vxx+vyy=0+0=0. The polynomial components are C2, so [L3] identifies them as harmonic and identifies v as a harmonic conjugate of u.

step 3.1step 3.2L3algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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