Metamath Proof Explorer


Theorem ringnegr

Description: Negation in a ring is the same as right multiplication by -1. ( rngonegmn1r analog.) (Contributed by Jeff Madsen, 19-Jun-2010) (Revised by Mario Carneiro, 2-Jul-2014)

Ref Expression
Hypotheses ringnegl.b B=BaseR
ringnegl.t ·˙=R
ringnegl.u 1˙=1R
ringnegl.n N=invgR
ringnegl.r φRRing
ringnegl.x φXB
Assertion ringnegr φX·˙N1˙=NX

Proof

Step Hyp Ref Expression
1 ringnegl.b B=BaseR
2 ringnegl.t ·˙=R
3 ringnegl.u 1˙=1R
4 ringnegl.n N=invgR
5 ringnegl.r φRRing
6 ringnegl.x φXB
7 ringgrp RRingRGrp
8 5 7 syl φRGrp
9 1 3 ringidcl RRing1˙B
10 5 9 syl φ1˙B
11 1 4 grpinvcl RGrp1˙BN1˙B
12 8 10 11 syl2anc φN1˙B
13 eqid +R=+R
14 1 13 2 ringdi RRingXBN1˙B1˙BX·˙N1˙+R1˙=X·˙N1˙+RX·˙1˙
15 5 6 12 10 14 syl13anc φX·˙N1˙+R1˙=X·˙N1˙+RX·˙1˙
16 eqid 0R=0R
17 1 13 16 4 grplinv RGrp1˙BN1˙+R1˙=0R
18 8 10 17 syl2anc φN1˙+R1˙=0R
19 18 oveq2d φX·˙N1˙+R1˙=X·˙0R
20 1 2 16 ringrz RRingXBX·˙0R=0R
21 5 6 20 syl2anc φX·˙0R=0R
22 19 21 eqtrd φX·˙N1˙+R1˙=0R
23 1 2 3 ringridm RRingXBX·˙1˙=X
24 5 6 23 syl2anc φX·˙1˙=X
25 24 oveq2d φX·˙N1˙+RX·˙1˙=X·˙N1˙+RX
26 15 22 25 3eqtr3rd φX·˙N1˙+RX=0R
27 1 2 ringcl RRingXBN1˙BX·˙N1˙B
28 5 6 12 27 syl3anc φX·˙N1˙B
29 1 13 16 4 grpinvid2 RGrpXBX·˙N1˙BNX=X·˙N1˙X·˙N1˙+RX=0R
30 8 6 28 29 syl3anc φNX=X·˙N1˙X·˙N1˙+RX=0R
31 26 30 mpbird φNX=X·˙N1˙
32 31 eqcomd φX·˙N1˙=NX