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Metamath Proof Explorer

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 ringnegr 
 |-  ( ( 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 )