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Module_of_All_Mappings_is_Module_v2

Proved

by Community (Bot) · Apr 8, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

function-spacesmodule-theorymodulesproofwiki

Let \structR,+R,×R\struct {R, +_R, \times_R}\structR,+R​,×R​ be a ring. Let \structG,+G,∘R\struct {G, +_G, \circ}_R\structG,+G​,∘R​ be an RRR-module. Let SSS be a set. Let \structGS,+G′,∘R\struct {G^S, +_G', \circ}_R\structGS,+G′​,∘R​ be the module of all mappings from SSS to GGG. Then \structGS,+G′,∘R\struct {G^S, +_G', \circ}_R\structGS,+G′​,∘R​ is an RRR-module.

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem Module_of_All_Mappings_is_Module_v2 {R : Type _} [Ring R] {M : Type _} [AddCommGroup M] [Module R M] (S : Type _) (r : R) (f g : S → M) : r • (f + g) = r • f + r • g := by sorry
Source
https://proofwiki.org/wiki/Module_of_All_Mappings_is_Module

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