Theorem ringidmlem

Description: Lemma for ringlidm and ringridm . (Contributed by NM, 15-Sep-2011) (Revised by Mario Carneiro, 27-Dec-2014)

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

Proof

Step	Hyp	Ref	Expression
1		rngidm.b	|- B = ( Base ` R )
2		rngidm.t	|- .x. = ( .r ` R )
3		rngidm.u	|- .1. = ( 1r ` R )
4		eqid	|- ( mulGrp ` R ) = ( mulGrp ` R )
5	4	ringmgp	|- ( R e. Ring -> ( 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. Ring /\ X e. B ) -> ( ( .1. .x. X ) = X /\ ( X .x. .1. ) = X ) )