Metamath Proof Explorer


Theorem rngonegmn1l

Description: Negation in a ring is the same as left multiplication by -u 1 . (Contributed by Jeff Madsen, 10-Jun-2010)

Ref Expression
Hypotheses ringneg.1 G = 1 st R
ringneg.2 H = 2 nd R
ringneg.3 X = ran G
ringneg.4 N = inv G
ringneg.5 U = GId H
Assertion rngonegmn1l R RingOps A X N A = N U H A

Proof

Step Hyp Ref Expression
1 ringneg.1 G = 1 st R
2 ringneg.2 H = 2 nd R
3 ringneg.3 X = ran G
4 ringneg.4 N = inv G
5 ringneg.5 U = GId H
6 1 rneqi ran G = ran 1 st R
7 3 6 eqtri X = ran 1 st R
8 7 2 5 rngo1cl R RingOps U X
9 1 3 4 rngonegcl R RingOps U X N U X
10 8 9 mpdan R RingOps N U X
11 8 10 jca R RingOps U X N U X
12 1 2 3 rngodir R RingOps U X N U X A X U G N U H A = U H A G N U H A
13 12 3exp2 R RingOps U X N U X A X U G N U H A = U H A G N U H A
14 13 imp42 R RingOps U X N U X A X U G N U H A = U H A G N U H A
15 14 an32s R RingOps A X U X N U X U G N U H A = U H A G N U H A
16 11 15 mpidan R RingOps A X U G N U H A = U H A G N U H A
17 eqid GId G = GId G
18 1 3 4 17 rngoaddneg1 R RingOps U X U G N U = GId G
19 8 18 mpdan R RingOps U G N U = GId G
20 19 adantr R RingOps A X U G N U = GId G
21 20 oveq1d R RingOps A X U G N U H A = GId G H A
22 17 3 1 2 rngolz R RingOps A X GId G H A = GId G
23 21 22 eqtrd R RingOps A X U G N U H A = GId G
24 2 7 5 rngolidm R RingOps A X U H A = A
25 24 oveq1d R RingOps A X U H A G N U H A = A G N U H A
26 16 23 25 3eqtr3rd R RingOps A X A G N U H A = GId G
27 1 2 3 rngocl R RingOps N U X A X N U H A X
28 27 3expa R RingOps N U X A X N U H A X
29 28 an32s R RingOps A X N U X N U H A X
30 10 29 mpidan R RingOps A X N U H A X
31 1 rngogrpo R RingOps G GrpOp
32 3 17 4 grpoinvid1 G GrpOp A X N U H A X N A = N U H A A G N U H A = GId G
33 31 32 syl3an1 R RingOps A X N U H A X N A = N U H A A G N U H A = GId G
34 30 33 mpd3an3 R RingOps A X N A = N U H A A G N U H A = GId G
35 26 34 mpbird R RingOps A X N A = N U H A