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Metamath Proof Explorer

Theorem srgidmlem

Description: Lemma for srglidm and srgridm . (Contributed by NM, 15-Sep-2011) (Revised by Mario Carneiro, 27-Dec-2014) (Revised by Thierry Arnoux, 1-Apr-2018)

Ref Expression
Hypotheses srgidm.b 
|- B = ( Base ` R )
srgidm.t 
|- .x. = ( .r ` R )
srgidm.u 
|- .1. = ( 1r ` R )
Assertion srgidmlem 
|- ( ( R e. SRing /\ X e. B ) -> ( ( .1. .x. X ) = X /\ ( X .x. .1. ) = X ) )

Proof

Step Hyp Ref Expression
1 srgidm.b 
 |-  B = ( Base ` R )
2 srgidm.t 
 |-  .x. = ( .r ` R )
3 srgidm.u 
 |-  .1. = ( 1r ` R )
4 eqid 
 |-  ( mulGrp ` R ) = ( mulGrp ` R )
5 4 srgmgp 
 |-  ( R e. SRing -> ( mulGrp ` R ) e. Mnd )
6 4 1 mgpbas 
 |-  B = ( Base ` ( mulGrp ` R ) )
7 4 2 mgpplusg 
 |-  .x. = ( +g ` ( mulGrp ` R ) )
8 4 3 ringidval 
 |-  .1. = ( 0g ` ( mulGrp ` R ) )
9 6 7 8 mndlrid 
 |-  ( ( ( mulGrp ` R ) e. Mnd /\ X e. B ) -> ( ( .1. .x. X ) = X /\ ( X .x. .1. ) = X ) )
10 5 9 sylan 
 |-  ( ( R e. SRing /\ X e. B ) -> ( ( .1. .x. X ) = X /\ ( X .x. .1. ) = X ) )