Metamath Proof Explorer


Theorem poprelb

Description: Equality for unordered pairs with partially ordered elements. (Contributed by AV, 9-Jul-2023)

Ref Expression
Assertion poprelb ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A B = C D ↔ A = C ∧ B = D

Proof

Step Hyp Ref Expression
1 simp2 ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A ∈ X ∧ B ∈ X
2 an3 ⊢ Rel ⁡ R ∧ R Po X ∧ A R B ∧ C R D → Rel ⁡ R ∧ C R D
3 2 3adant2 ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → Rel ⁡ R ∧ C R D
4 brrelex12 ⊢ Rel ⁡ R ∧ C R D → C ∈ V ∧ D ∈ V
5 3 4 syl ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → C ∈ V ∧ D ∈ V
6 preq12bg ⊢ A ∈ X ∧ B ∈ X ∧ C ∈ V ∧ D ∈ V → A B = C D ↔ A = C ∧ B = D ∨ A = D ∧ B = C
7 1 5 6 syl2anc ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A B = C D ↔ A = C ∧ B = D ∨ A = D ∧ B = C
8 idd ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A = C ∧ B = D → A = C ∧ B = D
9 breq12 ⊢ B = C ∧ A = D → B R A ↔ C R D
10 9 ancoms ⊢ A = D ∧ B = C → B R A ↔ C R D
11 10 bicomd ⊢ A = D ∧ B = C → C R D ↔ B R A
12 11 anbi2d ⊢ A = D ∧ B = C → A R B ∧ C R D ↔ A R B ∧ B R A
13 po2nr ⊢ R Po X ∧ A ∈ X ∧ B ∈ X → ¬ A R B ∧ B R A
14 13 adantll ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X → ¬ A R B ∧ B R A
15 14 pm2.21d ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X → A R B ∧ B R A → A = C ∧ B = D
16 15 ex ⊢ Rel ⁡ R ∧ R Po X → A ∈ X ∧ B ∈ X → A R B ∧ B R A → A = C ∧ B = D
17 16 com13 ⊢ A R B ∧ B R A → A ∈ X ∧ B ∈ X → Rel ⁡ R ∧ R Po X → A = C ∧ B = D
18 12 17 biimtrdi ⊢ A = D ∧ B = C → A R B ∧ C R D → A ∈ X ∧ B ∈ X → Rel ⁡ R ∧ R Po X → A = C ∧ B = D
19 18 com23 ⊢ A = D ∧ B = C → A ∈ X ∧ B ∈ X → A R B ∧ C R D → Rel ⁡ R ∧ R Po X → A = C ∧ B = D
20 19 com14 ⊢ Rel ⁡ R ∧ R Po X → A ∈ X ∧ B ∈ X → A R B ∧ C R D → A = D ∧ B = C → A = C ∧ B = D
21 20 3imp ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A = D ∧ B = C → A = C ∧ B = D
22 8 21 jaod ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A = C ∧ B = D ∨ A = D ∧ B = C → A = C ∧ B = D
23 orc ⊢ A = C ∧ B = D → A = C ∧ B = D ∨ A = D ∧ B = C
24 22 23 impbid1 ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A = C ∧ B = D ∨ A = D ∧ B = C ↔ A = C ∧ B = D
25 7 24 bitrd ⊢ Rel ⁡ R ∧ R Po X ∧ A ∈ X ∧ B ∈ X ∧ A R B ∧ C R D → A B = C D ↔ A = C ∧ B = D