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Metamath Proof Explorer

Theorem isrim

Description: An isomorphism of rings is a bijective homomorphism. (Contributed by AV, 22-Oct-2019) Remove sethood antecedent. (Revised by SN, 12-Jan-2025)

Ref Expression
Hypotheses rhmf1o.b ⊢ B = Base R
rhmf1o.c ⊢ C = Base S
Assertion isrim ⊢ F ∈ R RingIso S ↔ F ∈ R RingHom S ∧ F : B ⟶ 1-1 onto C

Proof

Step Hyp Ref Expression
1 rhmf1o.b ⊢ B = Base R
2 rhmf1o.c ⊢ C = Base S
3 isrim0 ⊢ F ∈ R RingIso S ↔ F ∈ R RingHom S ∧ F -1 ∈ S RingHom R
4 1 2 rhmf1o ⊢ F ∈ R RingHom S → F : B ⟶ 1-1 onto C ↔ F -1 ∈ S RingHom R
5 4 bicomd ⊢ F ∈ R RingHom S → F -1 ∈ S RingHom R ↔ F : B ⟶ 1-1 onto C
6 5 pm5.32i ⊢ F ∈ R RingHom S ∧ F -1 ∈ S RingHom R ↔ F ∈ R RingHom S ∧ F : B ⟶ 1-1 onto C
7 3 6 bitri ⊢ F ∈ R RingIso S ↔ F ∈ R RingHom S ∧ F : B ⟶ 1-1 onto C