Theorem dvdsrneg

Description: An element divides its negative. (Contributed by Mario Carneiro, 1-Dec-2014)

	Ref	Expression
Hypotheses	dvdsr.1	|- B = ( Base ` R )
	dvdsr.2	|- .|| = ( ||r ` R )
	dvdsrneg.5	|- N = ( invg ` R )
Assertion	dvdsrneg	|- ( ( R e. Ring /\ X e. B ) -> X .|| ( N ` X ) )

Proof

Step	Hyp	Ref	Expression
1		dvdsr.1	|- B = ( Base ` R )
2		dvdsr.2	|- .|| = ( ||r ` R )
3		dvdsrneg.5	|- N = ( invg ` R )
4		id	|- ( X e. B -> X e. B )
5		ringgrp	|- ( R e. Ring -> R e. Grp )
6		eqid	|- ( 1r ` R ) = ( 1r ` R )
7	1 6	ringidcl	|- ( R e. Ring -> ( 1r ` R ) e. B )
8	1 3	grpinvcl	|- ( ( R e. Grp /\ ( 1r ` R ) e. B ) -> ( N ` ( 1r ` R ) ) e. B )
9	5 7 8	syl2anc	|- ( R e. Ring -> ( N ` ( 1r ` R ) ) e. B )
10		eqid	|- ( .r ` R ) = ( .r ` R )
11	1 2 10	dvdsrmul	|- ( ( X e. B /\ ( N ` ( 1r ` R ) ) e. B ) -> X .|| ( ( N ` ( 1r ` R ) ) ( .r ` R ) X ) )
12	4 9 11	syl2anr	|- ( ( R e. Ring /\ X e. B ) -> X .|| ( ( N ` ( 1r ` R ) ) ( .r ` R ) X ) )
13		simpl	|- ( ( R e. Ring /\ X e. B ) -> R e. Ring )
14		simpr	|- ( ( R e. Ring /\ X e. B ) -> X e. B )
15	1 10 6 3 13 14	ringnegl	|- ( ( R e. Ring /\ X e. B ) -> ( ( N ` ( 1r ` R ) ) ( .r ` R ) X ) = ( N ` X ) )
16	12 15	breqtrd	|- ( ( R e. Ring /\ X e. B ) -> X .|| ( N ` X ) )