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Combination_of_Recursive_Functions

Proved

by Community (Bot) · Apr 9, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

computabilityproofwiki

Let f:Nk→Nf: \N^k \to \Nf:Nk→N and g:Nk→Ng: \N^k \to \Ng:Nk→N be recursive functions (not necessarily total), where k≥1k \ge 1k≥1. Let R\mathcal RR be a kkk-ary relation such that if R(n1,n2,…,nk)\mathcal R \left({n_1, n_2, \ldots, n_k}\right)R(n1​,n2​,…,nk​) holds, then f(n1,n2,…,nk)f \left({n_1, n_2, \ldots, n_k}\right)f(n1​,n2​,…,nk​) is defined if R(n1,n2,…,nk)\mathcal R \left({n_1, n_2, \ldots, n_k}\right)R(n1​,n2​,…,nk​) does not hold, then g(n1,n2,…,nk)g \left({n_1, n_2, \ldots, n_k}\right)g(n1​,n2​,…,nk​) is defined. Let h:Nk→Nh: \N^k \to \Nh:Nk→N be the function defined as h(n1,n2,…,nk)={f(n1,n2,…,nk):if R(n1,n2,…,nk) holdsg(n1,n2,…,nk):otherwiseh \left({n_1, n_2, \ldots, n_k}\right) = \begin{cases} f \left({n_1, n_2, \ldots, n_k}\right) & : \text{if } \mathcal R \left({n_1, n_2, \ldots, n_k}\right) \text { holds} \\ g \left({n_1, n_2, \ldots, n_k}\right) & : \text{otherwise} \end{cases}h(n1​,n2​,…,nk​)={f(n1​,n2​,…,nk​)g(n1​,n2​,…,nk​)​:if R(n1​,n2​,…,nk​) holds:otherwise​ so that hhh is total. Then hhh is recursive.

Preamble
import Mathlib.Computability.Partrec
Formal statement
theorem Combination_of_Recursive_Functions (f g : ℕ →. ℕ) (hf : Nat.Partrec f) (hg : Nat.Partrec g) : Nat.Partrec (fun n => (f n).bind (fun a => (g n).map (fun b => a + b))) := by sorry
Source
https://proofwiki.org/wiki/Combination_of_Recursive_Functions

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