Metamath Proof Explorer


Theorem lmicrcl

Description: Isomorphism implies the right side is a module. (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Assertion lmicrcl ⊢ R ≃ 𝑚 S → S ∈ LMod

Proof

Step Hyp Ref Expression
1 brlmic ⊢ R ≃ 𝑚 S ↔ R LMIso S ≠ ∅
2 n0 ⊢ R LMIso S ≠ ∅ ↔ ∃ f f ∈ R LMIso S
3 1 2 bitri ⊢ R ≃ 𝑚 S ↔ ∃ f f ∈ R LMIso S
4 lmimlmhm ⊢ f ∈ R LMIso S → f ∈ R LMHom S
5 lmhmlmod2 ⊢ f ∈ R LMHom S → S ∈ LMod
6 4 5 syl ⊢ f ∈ R LMIso S → S ∈ LMod
7 6 exlimiv ⊢ ∃ f f ∈ R LMIso S → S ∈ LMod
8 3 7 sylbi ⊢ R ≃ 𝑚 S → S ∈ LMod