Metamath Proof Explorer


Theorem lmictra

Description: Module isomorphism is transitive. (Contributed by AV, 10-Mar-2019)

Ref Expression
Assertion lmictra ⊢ R ≃ 𝑚 S ∧ S ≃ 𝑚 T → R ≃ 𝑚 T

Proof

Step Hyp Ref Expression
1 brlmic ⊢ R ≃ 𝑚 S ↔ R LMIso S ≠ ∅
2 brlmic ⊢ S ≃ 𝑚 T ↔ S LMIso T ≠ ∅
3 n0 ⊢ R LMIso S ≠ ∅ ↔ ∃ g g ∈ R LMIso S
4 n0 ⊢ S LMIso T ≠ ∅ ↔ ∃ f f ∈ S LMIso T
5 lmimco ⊢ f ∈ S LMIso T ∧ g ∈ R LMIso S → f ∘ g ∈ R LMIso T
6 brlmici ⊢ f ∘ g ∈ R LMIso T → R ≃ 𝑚 T
7 5 6 syl ⊢ f ∈ S LMIso T ∧ g ∈ R LMIso S → R ≃ 𝑚 T
8 7 ex ⊢ f ∈ S LMIso T → g ∈ R LMIso S → R ≃ 𝑚 T
9 8 exlimiv ⊢ ∃ f f ∈ S LMIso T → g ∈ R LMIso S → R ≃ 𝑚 T
10 9 com12 ⊢ g ∈ R LMIso S → ∃ f f ∈ S LMIso T → R ≃ 𝑚 T
11 10 exlimiv ⊢ ∃ g g ∈ R LMIso S → ∃ f f ∈ S LMIso T → R ≃ 𝑚 T
12 11 imp ⊢ ∃ g g ∈ R LMIso S ∧ ∃ f f ∈ S LMIso T → R ≃ 𝑚 T
13 3 4 12 syl2anb ⊢ R LMIso S ≠ ∅ ∧ S LMIso T ≠ ∅ → R ≃ 𝑚 T
14 1 2 13 syl2anb ⊢ R ≃ 𝑚 S ∧ S ≃ 𝑚 T → R ≃ 𝑚 T