Theorem irredneg

Description: The negative of an irreducible element is irreducible. (Contributed by Mario Carneiro, 4-Dec-2014)

	Ref	Expression
Hypotheses	irredn0.i	|- I = ( Irred ` R )
	irredneg.n	|- N = ( invg ` R )
Assertion	irredneg	|- ( ( R e. Ring /\ X e. I ) -> ( N ` X ) e. I )

Proof

Step	Hyp	Ref	Expression
1		irredn0.i	|- I = ( Irred ` R )
2		irredneg.n	|- N = ( invg ` R )
3		eqid	|- ( Base ` R ) = ( Base ` R )
4		eqid	|- ( .r ` R ) = ( .r ` R )
5		eqid	|- ( 1r ` R ) = ( 1r ` R )
6		simpl	|- ( ( R e. Ring /\ X e. I ) -> R e. Ring )
7	1 3	irredcl	|- ( X e. I -> X e. ( Base ` R ) )
8	7	adantl	|- ( ( R e. Ring /\ X e. I ) -> X e. ( Base ` R ) )
9	3 4 5 2 6 8	rngnegr	|- ( ( R e. Ring /\ X e. I ) -> ( X ( .r ` R ) ( N ` ( 1r ` R ) ) ) = ( N ` X ) )
10		eqid	|- ( Unit ` R ) = ( Unit ` R )
11	10 5	1unit	|- ( R e. Ring -> ( 1r ` R ) e. ( Unit ` R ) )
12	10 2	unitnegcl	|- ( ( R e. Ring /\ ( 1r ` R ) e. ( Unit ` R ) ) -> ( N ` ( 1r ` R ) ) e. ( Unit ` R ) )
13	11 12	mpdan	|- ( R e. Ring -> ( N ` ( 1r ` R ) ) e. ( Unit ` R ) )
14	13	adantr	|- ( ( R e. Ring /\ X e. I ) -> ( N ` ( 1r ` R ) ) e. ( Unit ` R ) )
15	1 10 4	irredrmul	|- ( ( R e. Ring /\ X e. I /\ ( N ` ( 1r ` R ) ) e. ( Unit ` R ) ) -> ( X ( .r ` R ) ( N ` ( 1r ` R ) ) ) e. I )
16	14 15	mpd3an3	|- ( ( R e. Ring /\ X e. I ) -> ( X ( .r ` R ) ( N ` ( 1r ` R ) ) ) e. I )
17	9 16	eqeltrrd	|- ( ( R e. Ring /\ X e. I ) -> ( N ` X ) e. I )