Metamath Proof Explorer


Theorem linvh

Description: If an element has a unique left inverse, then the value satisfies the left inverse value equation. (Contributed by metakunt, 25-Apr-2025)

Ref Expression
Hypotheses linvh.1
|- ( ph -> X e. ( Base ` R ) )
linvh.2
|- ( ph -> E! i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) )
Assertion linvh
|- ( ph -> ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) = ( 0g ` R ) )

Proof

Step Hyp Ref Expression
1 linvh.1
 |-  ( ph -> X e. ( Base ` R ) )
2 linvh.2
 |-  ( ph -> E! i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) )
3 eqid
 |-  ( Base ` R ) = ( Base ` R )
4 eqid
 |-  ( +g ` R ) = ( +g ` R )
5 eqid
 |-  ( 0g ` R ) = ( 0g ` R )
6 eqid
 |-  ( invg ` R ) = ( invg ` R )
7 3 4 5 6 grpinvval
 |-  ( X e. ( Base ` R ) -> ( ( invg ` R ) ` X ) = ( iota_ i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) ) )
8 1 7 syl
 |-  ( ph -> ( ( invg ` R ) ` X ) = ( iota_ i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) ) )
9 riotacl2
 |-  ( E! i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) -> ( iota_ i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) ) e. { i e. ( Base ` R ) | ( i ( +g ` R ) X ) = ( 0g ` R ) } )
10 2 9 syl
 |-  ( ph -> ( iota_ i e. ( Base ` R ) ( i ( +g ` R ) X ) = ( 0g ` R ) ) e. { i e. ( Base ` R ) | ( i ( +g ` R ) X ) = ( 0g ` R ) } )
11 8 10 eqeltrd
 |-  ( ph -> ( ( invg ` R ) ` X ) e. { i e. ( Base ` R ) | ( i ( +g ` R ) X ) = ( 0g ` R ) } )
12 oveq1
 |-  ( i = ( ( invg ` R ) ` X ) -> ( i ( +g ` R ) X ) = ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) )
13 12 eqeq1d
 |-  ( i = ( ( invg ` R ) ` X ) -> ( ( i ( +g ` R ) X ) = ( 0g ` R ) <-> ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) = ( 0g ` R ) ) )
14 13 elrab
 |-  ( ( ( invg ` R ) ` X ) e. { i e. ( Base ` R ) | ( i ( +g ` R ) X ) = ( 0g ` R ) } <-> ( ( ( invg ` R ) ` X ) e. ( Base ` R ) /\ ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) = ( 0g ` R ) ) )
15 14 simprbi
 |-  ( ( ( invg ` R ) ` X ) e. { i e. ( Base ` R ) | ( i ( +g ` R ) X ) = ( 0g ` R ) } -> ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) = ( 0g ` R ) )
16 11 15 syl
 |-  ( ph -> ( ( ( invg ` R ) ` X ) ( +g ` R ) X ) = ( 0g ` R ) )