Metamath Proof Explorer


Theorem grpsubid

Description: Subtraction of a group element from itself. (Contributed by NM, 31-Mar-2014)

Ref Expression
Hypotheses grpsubid.b
|- B = ( Base ` G )
grpsubid.o
|- .0. = ( 0g ` G )
grpsubid.m
|- .- = ( -g ` G )
Assertion grpsubid
|- ( ( G e. Grp /\ X e. B ) -> ( X .- X ) = .0. )

Proof

Step Hyp Ref Expression
1 grpsubid.b
 |-  B = ( Base ` G )
2 grpsubid.o
 |-  .0. = ( 0g ` G )
3 grpsubid.m
 |-  .- = ( -g ` G )
4 eqid
 |-  ( +g ` G ) = ( +g ` G )
5 eqid
 |-  ( invg ` G ) = ( invg ` G )
6 1 4 5 3 grpsubval
 |-  ( ( X e. B /\ X e. B ) -> ( X .- X ) = ( X ( +g ` G ) ( ( invg ` G ) ` X ) ) )
7 6 anidms
 |-  ( X e. B -> ( X .- X ) = ( X ( +g ` G ) ( ( invg ` G ) ` X ) ) )
8 7 adantl
 |-  ( ( G e. Grp /\ X e. B ) -> ( X .- X ) = ( X ( +g ` G ) ( ( invg ` G ) ` X ) ) )
9 1 4 2 5 grprinv
 |-  ( ( G e. Grp /\ X e. B ) -> ( X ( +g ` G ) ( ( invg ` G ) ` X ) ) = .0. )
10 8 9 eqtrd
 |-  ( ( G e. Grp /\ X e. B ) -> ( X .- X ) = .0. )