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Metamath Proof Explorer

Theorem grpsubid1

Description: Subtraction of the identity from a group element. (Contributed by Mario Carneiro, 14-Jan-2015)

Ref Expression
Hypotheses grpsubid.b 
|- B = ( Base ` G )
grpsubid.o 
|- .0. = ( 0g ` G )
grpsubid.m 
|- .- = ( -g ` G )
Assertion grpsubid1 
|- ( ( G e. Grp /\ X e. B ) -> ( X .- .0. ) = X )

Proof

Step Hyp Ref Expression
1 grpsubid.b 
 |-  B = ( Base ` G )
2 grpsubid.o 
 |-  .0. = ( 0g ` G )
3 grpsubid.m 
 |-  .- = ( -g ` G )
4 id 
 |-  ( X e. B -> X e. B )
5 1 2 grpidcl 
 |-  ( G e. Grp -> .0. e. B )
6 eqid 
 |-  ( +g ` G ) = ( +g ` G )
7 eqid 
 |-  ( invg ` G ) = ( invg ` G )
8 1 6 7 3 grpsubval 
 |-  ( ( X e. B /\ .0. e. B ) -> ( X .- .0. ) = ( X ( +g ` G ) ( ( invg ` G ) ` .0. ) ) )
9 4 5 8 syl2anr 
 |-  ( ( G e. Grp /\ X e. B ) -> ( X .- .0. ) = ( X ( +g ` G ) ( ( invg ` G ) ` .0. ) ) )
10 2 7 grpinvid 
 |-  ( G e. Grp -> ( ( invg ` G ) ` .0. ) = .0. )
11 10 adantr 
 |-  ( ( G e. Grp /\ X e. B ) -> ( ( invg ` G ) ` .0. ) = .0. )
12 11 oveq2d 
 |-  ( ( G e. Grp /\ X e. B ) -> ( X ( +g ` G ) ( ( invg ` G ) ` .0. ) ) = ( X ( +g ` G ) .0. ) )
13 1 6 2 grprid 
 |-  ( ( G e. Grp /\ X e. B ) -> ( X ( +g ` G ) .0. ) = X )
14 9 12 13 3eqtrd 
 |-  ( ( G e. Grp /\ X e. B ) -> ( X .- .0. ) = X )