Metamath Proof Explorer


Theorem erldi

Description: Main property of the ring localization equivalence 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
erldi.4 ⊢ φ → U ∼ ˙ V
Assertion erldi ⊢ φ → ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙

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 erldi.4 ⊢ φ → U ∼ ˙ V
8 eqid ⊢ B × S = B × S
9 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 ˙
10 1 4 5 6 8 9 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 ˙
11 2 10 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 ˙
12 simpl ⊢ a = U ∧ b = V → a = U
13 12 fveq2d ⊢ a = U ∧ b = V → 1 st ⁡ a = 1 st ⁡ U
14 simpr ⊢ a = U ∧ b = V → b = V
15 14 fveq2d ⊢ a = U ∧ b = V → 2 nd ⁡ b = 2 nd ⁡ V
16 13 15 oveq12d ⊢ a = U ∧ b = V → 1 st ⁡ a · ˙ 2 nd ⁡ b = 1 st ⁡ U · ˙ 2 nd ⁡ V
17 14 fveq2d ⊢ a = U ∧ b = V → 1 st ⁡ b = 1 st ⁡ V
18 12 fveq2d ⊢ a = U ∧ b = V → 2 nd ⁡ a = 2 nd ⁡ U
19 17 18 oveq12d ⊢ a = U ∧ b = V → 1 st ⁡ b · ˙ 2 nd ⁡ a = 1 st ⁡ V · ˙ 2 nd ⁡ U
20 16 19 oveq12d ⊢ a = U ∧ b = V → 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U
21 20 oveq2d ⊢ a = U ∧ b = V → t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U
22 21 eqeq1d ⊢ a = U ∧ b = V → t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ ↔ t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙
23 22 rexbidv ⊢ a = U ∧ b = V → ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ ↔ ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙
24 23 adantl ⊢ φ ∧ a = U ∧ b = V → ∃ t ∈ S t · ˙ 1 st ⁡ a · ˙ 2 nd ⁡ b - ˙ 1 st ⁡ b · ˙ 2 nd ⁡ a = 0 ˙ ↔ ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙
25 11 24 brab2d ⊢ φ → U ∼ ˙ V ↔ U ∈ B × S ∧ V ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙
26 7 25 mpbid ⊢ φ → U ∈ B × S ∧ V ∈ B × S ∧ ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙
27 26 simprd ⊢ φ → ∃ t ∈ S t · ˙ 1 st ⁡ U · ˙ 2 nd ⁡ V - ˙ 1 st ⁡ V · ˙ 2 nd ⁡ U = 0 ˙