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