Metamath Proof Explorer


Theorem opbrop

Description: Ordered pair membership in a relation. Special case. (Contributed by NM, 5-Aug-1995)

Ref Expression
Hypotheses opbrop.1 ⊢ z = A ∧ w = B ∧ v = C ∧ u = D → φ ↔ ψ
opbrop.2 ⊢ R = x y | x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ φ
Assertion opbrop ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A B R C D ↔ ψ

Proof

Step Hyp Ref Expression
1 opbrop.1 ⊢ z = A ∧ w = B ∧ v = C ∧ u = D → φ ↔ ψ
2 opbrop.2 ⊢ R = x y | x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ φ
3 opelxpi ⊢ A ∈ S ∧ B ∈ S → A B ∈ S × S
4 opelxpi ⊢ C ∈ S ∧ D ∈ S → C D ∈ S × S
5 3 4 anim12i ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A B ∈ S × S ∧ C D ∈ S × S
6 opex ⊢ A B ∈ V
7 opex ⊢ C D ∈ V
8 eleq1 ⊢ x = A B → x ∈ S × S ↔ A B ∈ S × S
9 8 anbi1d ⊢ x = A B → x ∈ S × S ∧ y ∈ S × S ↔ A B ∈ S × S ∧ y ∈ S × S
10 eqeq1 ⊢ x = A B → x = z w ↔ A B = z w
11 10 anbi1d ⊢ x = A B → x = z w ∧ y = v u ↔ A B = z w ∧ y = v u
12 11 anbi1d ⊢ x = A B → x = z w ∧ y = v u ∧ φ ↔ A B = z w ∧ y = v u ∧ φ
13 12 4exbidv ⊢ x = A B → ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ φ ↔ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ y = v u ∧ φ
14 9 13 anbi12d ⊢ x = A B → x ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u x = z w ∧ y = v u ∧ φ ↔ A B ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ y = v u ∧ φ
15 eleq1 ⊢ y = C D → y ∈ S × S ↔ C D ∈ S × S
16 15 anbi2d ⊢ y = C D → A B ∈ S × S ∧ y ∈ S × S ↔ A B ∈ S × S ∧ C D ∈ S × S
17 eqeq1 ⊢ y = C D → y = v u ↔ C D = v u
18 17 anbi2d ⊢ y = C D → A B = z w ∧ y = v u ↔ A B = z w ∧ C D = v u
19 18 anbi1d ⊢ y = C D → A B = z w ∧ y = v u ∧ φ ↔ A B = z w ∧ C D = v u ∧ φ
20 19 4exbidv ⊢ y = C D → ∃ z ∃ w ∃ v ∃ u A B = z w ∧ y = v u ∧ φ ↔ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ C D = v u ∧ φ
21 16 20 anbi12d ⊢ y = C D → A B ∈ S × S ∧ y ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ y = v u ∧ φ ↔ A B ∈ S × S ∧ C D ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ C D = v u ∧ φ
22 6 7 14 21 2 brab ⊢ A B R C D ↔ A B ∈ S × S ∧ C D ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ C D = v u ∧ φ
23 1 copsex4g ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → ∃ z ∃ w ∃ v ∃ u A B = z w ∧ C D = v u ∧ φ ↔ ψ
24 23 anbi2d ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A B ∈ S × S ∧ C D ∈ S × S ∧ ∃ z ∃ w ∃ v ∃ u A B = z w ∧ C D = v u ∧ φ ↔ A B ∈ S × S ∧ C D ∈ S × S ∧ ψ
25 22 24 bitrid ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A B R C D ↔ A B ∈ S × S ∧ C D ∈ S × S ∧ ψ
26 5 25 mpbirand ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A B R C D ↔ ψ