Metamath Proof Explorer


Theorem gicsym

Description: Isomorphism is symmetric. (Contributed by Mario Carneiro, 21-Apr-2016)

Ref Expression
Assertion gicsym
|- ( R ~=g S -> S ~=g R )

Proof

Step Hyp Ref Expression
1 brgic
 |-  ( R ~=g S <-> ( R GrpIso S ) =/= (/) )
2 n0
 |-  ( ( R GrpIso S ) =/= (/) <-> E. f f e. ( R GrpIso S ) )
3 gimcnv
 |-  ( f e. ( R GrpIso S ) -> `' f e. ( S GrpIso R ) )
4 brgici
 |-  ( `' f e. ( S GrpIso R ) -> S ~=g R )
5 3 4 syl
 |-  ( f e. ( R GrpIso S ) -> S ~=g R )
6 5 exlimiv
 |-  ( E. f f e. ( R GrpIso S ) -> S ~=g R )
7 2 6 sylbi
 |-  ( ( R GrpIso S ) =/= (/) -> S ~=g R )
8 1 7 sylbi
 |-  ( R ~=g S -> S ~=g R )