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 = Base R
ringnegl.t ⊢ · ˙ = ⋅ R
ringnegl.u ⊢ 1 ˙ = 1 R
ringnegl.n ⊢ N = inv g ⁡ R
ringnegl.r ⊢ φ → R ∈ Ring
ringnegl.x ⊢ φ → X ∈ B
Assertion ringnegr ⊢ φ → X · ˙ N ⁡ 1 ˙ = N ⁡ X

Proof

Step Hyp Ref Expression
1 ringnegl.b ⊢ B = Base R
2 ringnegl.t ⊢ · ˙ = ⋅ R
3 ringnegl.u ⊢ 1 ˙ = 1 R
4 ringnegl.n ⊢ N = inv g ⁡ R
5 ringnegl.r ⊢ φ → R ∈ Ring
6 ringnegl.x ⊢ φ → X ∈ B
7 ringgrp ⊢ R ∈ Ring → R ∈ Grp
8 5 7 syl ⊢ φ → R ∈ Grp
9 1 3 ringidcl ⊢ R ∈ Ring → 1 ˙ ∈ B
10 5 9 syl ⊢ φ → 1 ˙ ∈ B
11 1 4 grpinvcl ⊢ R ∈ Grp ∧ 1 ˙ ∈ B → N ⁡ 1 ˙ ∈ B
12 8 10 11 syl2anc ⊢ φ → N ⁡ 1 ˙ ∈ B
13 eqid ⊢ + R = + R
14 1 13 2 ringdi ⊢ R ∈ Ring ∧ X ∈ B ∧ N ⁡ 1 ˙ ∈ B ∧ 1 ˙ ∈ B → X · ˙ N ⁡ 1 ˙ + R 1 ˙ = X · ˙ N ⁡ 1 ˙ + R X · ˙ 1 ˙
15 5 6 12 10 14 syl13anc ⊢ φ → X · ˙ N ⁡ 1 ˙ + R 1 ˙ = X · ˙ N ⁡ 1 ˙ + R X · ˙ 1 ˙
16 eqid ⊢ 0 R = 0 R
17 1 13 16 4 grplinv ⊢ R ∈ Grp ∧ 1 ˙ ∈ B → N ⁡ 1 ˙ + R 1 ˙ = 0 R
18 8 10 17 syl2anc ⊢ φ → N ⁡ 1 ˙ + R 1 ˙ = 0 R
19 18 oveq2d ⊢ φ → X · ˙ N ⁡ 1 ˙ + R 1 ˙ = X · ˙ 0 R
20 1 2 16 ringrz ⊢ R ∈ Ring ∧ X ∈ B → X · ˙ 0 R = 0 R
21 5 6 20 syl2anc ⊢ φ → X · ˙ 0 R = 0 R
22 19 21 eqtrd ⊢ φ → X · ˙ N ⁡ 1 ˙ + R 1 ˙ = 0 R
23 1 2 3 ringridm ⊢ R ∈ Ring ∧ X ∈ B → X · ˙ 1 ˙ = X
24 5 6 23 syl2anc ⊢ φ → X · ˙ 1 ˙ = X
25 24 oveq2d ⊢ φ → X · ˙ N ⁡ 1 ˙ + R X · ˙ 1 ˙ = X · ˙ N ⁡ 1 ˙ + R X
26 15 22 25 3eqtr3rd ⊢ φ → X · ˙ N ⁡ 1 ˙ + R X = 0 R
27 1 2 ringcl ⊢ R ∈ Ring ∧ X ∈ B ∧ N ⁡ 1 ˙ ∈ B → X · ˙ N ⁡ 1 ˙ ∈ B
28 5 6 12 27 syl3anc ⊢ φ → X · ˙ N ⁡ 1 ˙ ∈ B
29 1 13 16 4 grpinvid2 ⊢ R ∈ Grp ∧ X ∈ B ∧ X · ˙ N ⁡ 1 ˙ ∈ B → N ⁡ X = X · ˙ N ⁡ 1 ˙ ↔ X · ˙ N ⁡ 1 ˙ + R X = 0 R
30 8 6 28 29 syl3anc ⊢ φ → N ⁡ X = X · ˙ N ⁡ 1 ˙ ↔ X · ˙ N ⁡ 1 ˙ + R X = 0 R
31 26 30 mpbird ⊢ φ → N ⁡ X = X · ˙ N ⁡ 1 ˙
32 31 eqcomd ⊢ φ → X · ˙ N ⁡ 1 ˙ = N ⁡ X