Metamath Proof Explorer

Theorem ringmneg2

Description: Negation of a product in a ring. ( mulneg2 analog.) (Contributed by Jeff Madsen, 19-Jun-2010) (Revised by Mario Carneiro, 2-Jul-2014)

Ref Expression
Hypotheses ringneglmul.b 
|- B = ( Base ` R )
ringneglmul.t 
|- .x. = ( .r ` R )
ringneglmul.n 
|- N = ( invg ` R )
ringneglmul.r 
|- ( ph -> R e. Ring )
ringneglmul.x 
|- ( ph -> X e. B )
ringneglmul.y 
|- ( ph -> Y e. B )
Assertion ringmneg2 
|- ( ph -> ( X .x. ( N ` Y ) ) = ( N ` ( X .x. Y ) ) )

Proof

Step Hyp Ref Expression
1 ringneglmul.b 
 |-  B = ( Base ` R )
2 ringneglmul.t 
 |-  .x. = ( .r ` R )
3 ringneglmul.n 
 |-  N = ( invg ` R )
4 ringneglmul.r 
 |-  ( ph -> R e. Ring )
5 ringneglmul.x 
 |-  ( ph -> X e. B )
6 ringneglmul.y 
 |-  ( ph -> Y e. B )
7 ringgrp 
 |-  ( R e. Ring -> R e. Grp )
8 4 7 syl 
 |-  ( ph -> R e. Grp )
9 eqid 
 |-  ( 1r ` R ) = ( 1r ` R )
10 1 9 ringidcl 
 |-  ( R e. Ring -> ( 1r ` R ) e. B )
11 4 10 syl 
 |-  ( ph -> ( 1r ` R ) e. B )
12 1 3 grpinvcl 
 |-  ( ( R e. Grp /\ ( 1r ` R ) e. B ) -> ( N ` ( 1r ` R ) ) e. B )
13 8 11 12 syl2anc 
 |-  ( ph -> ( N ` ( 1r ` R ) ) e. B )
14 1 2 ringass 
 |-  ( ( R e. Ring /\ ( X e. B /\ Y e. B /\ ( N ` ( 1r ` R ) ) e. B ) ) -> ( ( X .x. Y ) .x. ( N ` ( 1r ` R ) ) ) = ( X .x. ( Y .x. ( N ` ( 1r ` R ) ) ) ) )
15 4 5 6 13 14 syl13anc 
 |-  ( ph -> ( ( X .x. Y ) .x. ( N ` ( 1r ` R ) ) ) = ( X .x. ( Y .x. ( N ` ( 1r ` R ) ) ) ) )
16 1 2 ringcl 
 |-  ( ( R e. Ring /\ X e. B /\ Y e. B ) -> ( X .x. Y ) e. B )
17 4 5 6 16 syl3anc 
 |-  ( ph -> ( X .x. Y ) e. B )
18 1 2 9 3 4 17 rngnegr 
 |-  ( ph -> ( ( X .x. Y ) .x. ( N ` ( 1r ` R ) ) ) = ( N ` ( X .x. Y ) ) )
19 1 2 9 3 4 6 rngnegr 
 |-  ( ph -> ( Y .x. ( N ` ( 1r ` R ) ) ) = ( N ` Y ) )
20 19 oveq2d 
 |-  ( ph -> ( X .x. ( Y .x. ( N ` ( 1r ` R ) ) ) ) = ( X .x. ( N ` Y ) ) )
21 15 18 20 3eqtr3rd 
 |-  ( ph -> ( X .x. ( N ` Y ) ) = ( N ` ( X .x. Y ) ) )