Metamath Proof Explorer


Theorem invsym2

Description: The inverse relation is symmetric. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses invfval.b
|- B = ( Base ` C )
invfval.n
|- N = ( Inv ` C )
invfval.c
|- ( ph -> C e. Cat )
invfval.x
|- ( ph -> X e. B )
invfval.y
|- ( ph -> Y e. B )
Assertion invsym2
|- ( ph -> `' ( X N Y ) = ( Y N X ) )

Proof

Step Hyp Ref Expression
1 invfval.b
 |-  B = ( Base ` C )
2 invfval.n
 |-  N = ( Inv ` C )
3 invfval.c
 |-  ( ph -> C e. Cat )
4 invfval.x
 |-  ( ph -> X e. B )
5 invfval.y
 |-  ( ph -> Y e. B )
6 eqid
 |-  ( Hom ` C ) = ( Hom ` C )
7 1 2 3 5 4 6 invss
 |-  ( ph -> ( Y N X ) C_ ( ( Y ( Hom ` C ) X ) X. ( X ( Hom ` C ) Y ) ) )
8 relxp
 |-  Rel ( ( Y ( Hom ` C ) X ) X. ( X ( Hom ` C ) Y ) )
9 relss
 |-  ( ( Y N X ) C_ ( ( Y ( Hom ` C ) X ) X. ( X ( Hom ` C ) Y ) ) -> ( Rel ( ( Y ( Hom ` C ) X ) X. ( X ( Hom ` C ) Y ) ) -> Rel ( Y N X ) ) )
10 7 8 9 mpisyl
 |-  ( ph -> Rel ( Y N X ) )
11 relcnv
 |-  Rel `' ( X N Y )
12 10 11 jctil
 |-  ( ph -> ( Rel `' ( X N Y ) /\ Rel ( Y N X ) ) )
13 1 2 3 4 5 invsym
 |-  ( ph -> ( f ( X N Y ) g <-> g ( Y N X ) f ) )
14 vex
 |-  g e. _V
15 vex
 |-  f e. _V
16 14 15 brcnv
 |-  ( g `' ( X N Y ) f <-> f ( X N Y ) g )
17 df-br
 |-  ( g `' ( X N Y ) f <-> <. g , f >. e. `' ( X N Y ) )
18 16 17 bitr3i
 |-  ( f ( X N Y ) g <-> <. g , f >. e. `' ( X N Y ) )
19 df-br
 |-  ( g ( Y N X ) f <-> <. g , f >. e. ( Y N X ) )
20 13 18 19 3bitr3g
 |-  ( ph -> ( <. g , f >. e. `' ( X N Y ) <-> <. g , f >. e. ( Y N X ) ) )
21 20 eqrelrdv2
 |-  ( ( ( Rel `' ( X N Y ) /\ Rel ( Y N X ) ) /\ ph ) -> `' ( X N Y ) = ( Y N X ) )
22 12 21 mpancom
 |-  ( ph -> `' ( X N Y ) = ( Y N X ) )