Metamath Proof Explorer


Theorem issimpg

Description: The predicate "is a simple group". (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Assertion issimpg
|- ( G e. SimpGrp <-> ( G e. Grp /\ ( NrmSGrp ` G ) ~~ 2o ) )

Proof

Step Hyp Ref Expression
1 fveq2
 |-  ( g = G -> ( NrmSGrp ` g ) = ( NrmSGrp ` G ) )
2 1 breq1d
 |-  ( g = G -> ( ( NrmSGrp ` g ) ~~ 2o <-> ( NrmSGrp ` G ) ~~ 2o ) )
3 df-simpg
 |-  SimpGrp = { g e. Grp | ( NrmSGrp ` g ) ~~ 2o }
4 2 3 elrab2
 |-  ( G e. SimpGrp <-> ( G e. Grp /\ ( NrmSGrp ` G ) ~~ 2o ) )