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Liouville's theorem (differential algebra)

Proved
LiouvilleDiffAlg.liouville_basic_theorem

by Lucas · Sep 27, 2026 · Mathlib 0df444a (Lean v4.33.1)

differential-algebrasymbolic-integration

Let F⊆GF \subseteq GF⊆G be differential fields of characteristic zero with the same constants, Con⁡(F)=Con⁡(G)\operatorname{Con}(F) = \operatorname{Con}(G)Con(F)=Con(G), and suppose that GGG is an elementary differential extension of FFF. Suppose f∈Ff \in Ff∈F and g∈Gg \in Gg∈G satisfy Dg=fDg = fDg=f (that is, GGG contains an antiderivative of fff). Then there exist n≥0n \ge 0n≥0, constants c1,…,cn∈Con⁡(F)c_1, \dots, c_n \in \operatorname{Con}(F)c1​,…,cn​∈Con(F), nonzero elements f1,…,fn∈Ff_1, \dots, f_n \in Ff1​,…,fn​∈F and s∈Fs \in Fs∈F such that

f=c1Df1f1+⋯+cnDfnfn+Ds.f = c_1\frac{Df_1}{f_1} + \cdots + c_n\frac{Df_n}{f_n} + Ds.f=c1​f1​Df1​​+⋯+cn​fn​Dfn​​+Ds.

In words: if fff has an antiderivative in an elementary extension of FFF, then that antiderivative is an element of FFF plus a constant linear combination of logarithms of elements of FFF. This is the theorem on which the Risch algorithm is based.

Formalization Note The characteristic-zero hypothesis is not written in the source article. It is the standing convention of the theorem in its standard references (Rosenlicht 1972; Geddes–Czapor–Labahn §12.4). The equality of constants is stated as: the image of Con⁡(F)\operatorname{Con}(F)Con(F) in GGG equals Con⁡(G)\operatorname{Con}(G)Con(G).

Preamble
import Mathlib
import Definitions.Def_LiouvilleDiffAlg_Basic

open scoped Differential
Formal statement
namespace LiouvilleDiffAlg

theorem liouville_basic_theorem {F G : Type*} [Field F] [Field G] [Differential F]
    [Differential G] [Algebra F G] [DifferentialAlgebra F G] [CharZero F]
    (hcon : algebraMap F G '' constants F = constants G)
    (helem : IsElementaryDifferentialExtension F G)
    (f : F) (g : G) (hg : g′ = algebraMap F G f) :
    ∃ (n : ℕ) (c : Fin n → F) (u : Fin n → F) (v : F),
      (∀ i, c i ∈ constants F) ∧ (∀ i, u i ≠ 0) ∧
      f = ∑ i, c i * ((u i)′ / u i) + v′ := by sorry

end LiouvilleDiffAlg
Source
Wikipedia, "Liouville's theorem (differential algebra)", revision oldid=1349223559, https://en.wikipedia.org/w/index.php?title=Liouville%27s_theorem_(differential_algebra)&oldid=1349223559, section "Basic theorem"
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

Non-blind read-back — not independent testimony. This read-back was written by the same agent that drafted the Lean statements below (Aristotle, by Harmonic), with full knowledge of the source article and of the intended meaning. It was not produced by a blind, independent auditor, so it must not be mistaken for independent testimony; please compare it against the Lean code yourself.

Let FFF and GGG be fields, each with a derivation DDD (over Z\mathbb{Z}Z), with an algebra map ι:F→G\iota : F \to Gι:F→G commuting with the derivations (D ι(a)=ι(Da)D\,\iota(a) = \iota(Da)Dι(a)=ι(Da)), and assume FFF has characteristic zero. Assume:

  1. ι({a∈F:Da=0})={b∈G:Db=0}\iota(\{a \in F : Da = 0\}) = \{b \in G : Db = 0\}ι({a∈F:Da=0})={b∈G:Db=0} (equality of sets);
  2. GGG is an elementary differential extension of FFF: there are m∈Nm \in \mathbb{N}m∈N and intermediate fields K0=ι(F)⊆…K_0 = \iota(F) \subseteq \dotsK0​=ι(F)⊆…, Km=GK_m = GKm​=G, where each Ki+1K_{i+1}Ki+1​ (i<mi < mi<m) is generated over FFF by KiK_iKi​ and a single t∈Gt \in Gt∈G. This ttt is algebraic over KiK_iKi​, or transcendental over KiK_iKi​ with Dt=Ds/sDt = Ds/sDt=Ds/s for some nonzero s∈Kis \in K_is∈Ki​, or transcendental over KiK_iKi​ with Dt/t=DsDt/t = DsDt/t=Ds for some s∈Kis \in K_is∈Ki​, with derivatives computed in GGG;
  3. f∈Ff \in Ff∈F and g∈Gg \in Gg∈G with Dg=ι(f)Dg = \iota(f)Dg=ι(f).

Then there exist n∈Nn \in \mathbb{N}n∈N (possibly 000), elements c1,…,cn∈Fc_1, \dots, c_n \in Fc1​,…,cn​∈F with Dcj=0Dc_j = 0Dcj​=0, elements u1,…,un∈Fu_1, \dots, u_n \in Fu1​,…,un​∈F with every uj≠0u_j \neq 0uj​=0, and v∈Fv \in Fv∈F such that, in FFF,

f=∑j=1ncjDujuj+Dv.f = \sum_{j=1}^{n} c_j \frac{Du_j}{u_j} + Dv.f=j=1∑n​cj​uj​Duj​​+Dv.

When n=0n = 0n=0 the sum is empty and the conclusion says f=Dvf = Dvf=Dv.

Human review
  • Endorsed by Shuze Chen · Oct 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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