Metamath Proof Explorer


Theorem lmiclcl

Description: Isomorphism implies the left side is a module. (Contributed by Stefan O'Rear, 25-Jan-2015)

Ref Expression
Assertion lmiclcl ⊢ R ≃ 𝑚 S → R ∈ 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 lmhmlmod1 ⊢ f ∈ R LMHom S → R ∈ LMod
6 4 5 syl ⊢ f ∈ R LMIso S → R ∈ LMod
7 6 exlimiv ⊢ ∃ f f ∈ R LMIso S → R ∈ LMod
8 3 7 sylbi ⊢ R ≃ 𝑚 S → R ∈ LMod