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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. )