Metamath Proof Explorer


Theorem erlbrd

Description: Deduce the ring localization equivalence relation. If for some T e. S we have T x. ( E x. H - F x. G ) = 0 , then pairs <. E , G >. and <. F , H >. are equivalent under the localization relation. (Contributed by Thierry Arnoux, 4-May-2025)

Ref Expression
Hypotheses erlcl1.b ⊢ B = Base R
erlcl1.e ⊢ ∼ ˙ = R ~RL S
erlcl1.s ⊢ φ → S ⊆ B
erldi.1 ⊢ 0 ˙ = 0 R
erldi.2 ⊢ · ˙ = ⋅ R
erldi.3 ⊢ - ˙ = - R
erlbrd.u ⊢ φ → U = E G
erlbrd.v ⊢ φ → V = F H
erlbrd.e ⊢ φ → E ∈ B
erlbrd.f ⊢ φ → F ∈ B
erlbrd.g ⊢ φ → G ∈ S
erlbrd.h ⊢ φ → H ∈ S
erlbrd.1 ⊢ φ → T ∈ S
erlbrd.2 ⊢ φ → T · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
Assertion erlbrd ⊢ φ → U ∼ ˙ V

Proof

Step Hyp Ref Expression
1 erlcl1.b ⊢ B = Base R
2 erlcl1.e ⊢ ∼ ˙ = R ~RL S
3 erlcl1.s ⊢ φ → S ⊆ B
4 erldi.1 ⊢ 0 ˙ = 0 R
5 erldi.2 ⊢ · ˙ = ⋅ R
6 erldi.3 ⊢ - ˙ = - R
7 erlbrd.u ⊢ φ → U = E G
8 erlbrd.v ⊢ φ → V = F H
9 erlbrd.e ⊢ φ → E ∈ B
10 erlbrd.f ⊢ φ → F ∈ B
11 erlbrd.g ⊢ φ → G ∈ S
12 erlbrd.h ⊢ φ → H ∈ S
13 erlbrd.1 ⊢ φ → T ∈ S
14 erlbrd.2 ⊢ φ → T · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
15 9 11 opelxpd ⊢ φ → E G ∈ B × S
16 7 15 eqeltrd ⊢ φ → U ∈ B × S
17 10 12 opelxpd ⊢ φ → F H ∈ B × S
18 8 17 eqeltrd ⊢ φ → V ∈ B × S
19 16 18 jca ⊢ φ → U ∈ B × S ∧ V ∈ B × S
20 simpr ⊢ φ ∧ t = T → t = T
21 20 oveq1d ⊢ φ ∧ t = T → t · ˙ E · ˙ H - ˙ F · ˙ G = T · ˙ E · ˙ H - ˙ F · ˙ G
22 21 eqeq1d ⊢ φ ∧ t = T → t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙ ↔ T · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
23 13 22 14 rspcedvd ⊢ φ → ∃ t ∈ S t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
24 19 23 jca ⊢ φ → U ∈ B × S ∧ V ∈ B × S ∧ ∃ t ∈ S t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
25 eqid ⊢ B × S = B × S
26 eqid ⊢ a b | a ∈ B × S ∧ b ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ = a b | a ∈ B × S ∧ b ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙
27 1 4 5 6 25 26 3 erlval ⊢ φ → R ~RL S = a b | a ∈ B × S ∧ b ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙
28 2 27 eqtrid ⊢ φ → ∼ ˙ = a b | a ∈ B × S ∧ b ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙
29 simprl ⊢ φ ∧ a = U ∧ b = V → a = U
30 29 fveq2d ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ a = 1 st ⁡ U
31 7 fveq2d ⊢ φ → 1 st ⁡ U = 1 st ⁡ E G
32 op1stg ⊢ E ∈ B ∧ G ∈ S → 1 st ⁡ E G = E
33 9 11 32 syl2anc ⊢ φ → 1 st ⁡ E G = E
34 31 33 eqtrd ⊢ φ → 1 st ⁡ U = E
35 34 adantr ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ U = E
36 30 35 eqtrd ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ a = E
37 simprr ⊢ φ ∧ a = U ∧ b = V → b = V
38 37 fveq2d ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ b = 2 nd ⁡ V
39 8 fveq2d ⊢ φ → 2 nd ⁡ V = 2 nd ⁡ F H
40 op2ndg ⊢ F ∈ B ∧ H ∈ S → 2 nd ⁡ F H = H
41 10 12 40 syl2anc ⊢ φ → 2 nd ⁡ F H = H
42 39 41 eqtrd ⊢ φ → 2 nd ⁡ V = H
43 42 adantr ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ V = H
44 38 43 eqtrd ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ b = H
45 36 44 oveq12d ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ a · ˙ 2 nd ⁡ b = E · ˙ H
46 37 fveq2d ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ b = 1 st ⁡ V
47 8 fveq2d ⊢ φ → 1 st ⁡ V = 1 st ⁡ F H
48 op1stg ⊢ F ∈ B ∧ H ∈ S → 1 st ⁡ F H = F
49 10 12 48 syl2anc ⊢ φ → 1 st ⁡ F H = F
50 47 49 eqtrd ⊢ φ → 1 st ⁡ V = F
51 50 adantr ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ V = F
52 46 51 eqtrd ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ b = F
53 29 fveq2d ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ a = 2 nd ⁡ U
54 7 fveq2d ⊢ φ → 2 nd ⁡ U = 2 nd ⁡ E G
55 op2ndg ⊢ E ∈ B ∧ G ∈ S → 2 nd ⁡ E G = G
56 9 11 55 syl2anc ⊢ φ → 2 nd ⁡ E G = G
57 54 56 eqtrd ⊢ φ → 2 nd ⁡ U = G
58 57 adantr ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ U = G
59 53 58 eqtrd ⊢ φ ∧ a = U ∧ b = V → 2 nd ⁡ a = G
60 52 59 oveq12d ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ b · ˙ 2 nd ⁡ a = F · ˙ G
61 45 60 oveq12d ⊢ φ ∧ a = U ∧ b = V → 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = E · ˙ H - ˙ F · ˙ G
62 61 oveq2d ⊢ φ ∧ a = U ∧ b = V → t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = t · ˙ E · ˙ H - ˙ F · ˙ G
63 62 eqeq1d ⊢ φ ∧ a = U ∧ b = V → t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ ↔ t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
64 63 rexbidv ⊢ φ ∧ a = U ∧ b = V → ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ ↔ ∃ t ∈ S t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
65 28 64 brab2d ⊢ φ → U ∼ ˙ V ↔ U ∈ B × S ∧ V ∈ B × S ∧ ∃ t ∈ S t · ˙ E · ˙ H - ˙ F · ˙ G = 0 ˙
66 24 65 mpbird ⊢ φ → U ∼ ˙ V