Metamath Proof Explorer


Theorem isopropd

Description: Two structures with the same base, hom-sets and composition operation have the same isomorphisms. (Contributed by Zhi Wang, 27-Oct-2025)

Ref Expression
Hypotheses sectpropd.1
|- ( ph -> ( Homf ` C ) = ( Homf ` D ) )
sectpropd.2
|- ( ph -> ( comf ` C ) = ( comf ` D ) )
Assertion isopropd
|- ( ph -> ( Iso ` C ) = ( Iso ` D ) )

Proof

Step Hyp Ref Expression
1 sectpropd.1
 |-  ( ph -> ( Homf ` C ) = ( Homf ` D ) )
2 sectpropd.2
 |-  ( ph -> ( comf ` C ) = ( comf ` D ) )
3 1 2 isopropdlem
 |-  ( ( ph /\ f e. ( Iso ` C ) ) -> f e. ( Iso ` D ) )
4 1 eqcomd
 |-  ( ph -> ( Homf ` D ) = ( Homf ` C ) )
5 2 eqcomd
 |-  ( ph -> ( comf ` D ) = ( comf ` C ) )
6 4 5 isopropdlem
 |-  ( ( ph /\ f e. ( Iso ` D ) ) -> f e. ( Iso ` C ) )
7 3 6 impbida
 |-  ( ph -> ( f e. ( Iso ` C ) <-> f e. ( Iso ` D ) ) )
8 7 eqrdv
 |-  ( ph -> ( Iso ` C ) = ( Iso ` D ) )