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Metamath Proof Explorer

Theorem lmimco

Description: The composition of two isomorphisms of modules is an isomorphism of modules. (Contributed by AV, 10-Mar-2019)

Ref Expression
Assertion lmimco ⊢ F ∈ S LMIso T ∧ G ∈ R LMIso S → F ∘ G ∈ R LMIso T

Proof

Step Hyp Ref Expression
1 eqid ⊢ Base S = Base S
2 eqid ⊢ Base T = Base T
3 1 2 islmim ⊢ F ∈ S LMIso T ↔ F ∈ S LMHom T ∧ F : Base S ⟶ 1-1 onto Base T
4 eqid ⊢ Base R = Base R
5 4 1 islmim ⊢ G ∈ R LMIso S ↔ G ∈ R LMHom S ∧ G : Base R ⟶ 1-1 onto Base S
6 lmhmco ⊢ F ∈ S LMHom T ∧ G ∈ R LMHom S → F ∘ G ∈ R LMHom T
7 6 ad2ant2r ⊢ F ∈ S LMHom T ∧ F : Base S ⟶ 1-1 onto Base T ∧ G ∈ R LMHom S ∧ G : Base R ⟶ 1-1 onto Base S → F ∘ G ∈ R LMHom T
8 f1oco ⊢ F : Base S ⟶ 1-1 onto Base T ∧ G : Base R ⟶ 1-1 onto Base S → F ∘ G : Base R ⟶ 1-1 onto Base T
9 8 ad2ant2l ⊢ F ∈ S LMHom T ∧ F : Base S ⟶ 1-1 onto Base T ∧ G ∈ R LMHom S ∧ G : Base R ⟶ 1-1 onto Base S → F ∘ G : Base R ⟶ 1-1 onto Base T
10 4 2 islmim ⊢ F ∘ G ∈ R LMIso T ↔ F ∘ G ∈ R LMHom T ∧ F ∘ G : Base R ⟶ 1-1 onto Base T
11 7 9 10 sylanbrc ⊢ F ∈ S LMHom T ∧ F : Base S ⟶ 1-1 onto Base T ∧ G ∈ R LMHom S ∧ G : Base R ⟶ 1-1 onto Base S → F ∘ G ∈ R LMIso T
12 3 5 11 syl2anb ⊢ F ∈ S LMIso T ∧ G ∈ R LMIso S → F ∘ G ∈ R LMIso T