Metamath Proof Explorer


Theorem opelopab2a

Description: Ordered pair membership in an ordered pair class abstraction. (Contributed by Mario Carneiro, 19-Dec-2013)

Ref Expression
Hypothesis opelopabga.1 ⊢ x = A ∧ y = B → φ ↔ ψ
Assertion opelopab2a ⊢ A ∈ C ∧ B ∈ D → A B ∈ x y | x ∈ C ∧ y ∈ D ∧ φ ↔ ψ

Proof

Step Hyp Ref Expression
1 opelopabga.1 ⊢ x = A ∧ y = B → φ ↔ ψ
2 eleq1 ⊢ x = A → x ∈ C ↔ A ∈ C
3 eleq1 ⊢ y = B → y ∈ D ↔ B ∈ D
4 2 3 bi2anan9 ⊢ x = A ∧ y = B → x ∈ C ∧ y ∈ D ↔ A ∈ C ∧ B ∈ D
5 4 1 anbi12d ⊢ x = A ∧ y = B → x ∈ C ∧ y ∈ D ∧ φ ↔ A ∈ C ∧ B ∈ D ∧ ψ
6 5 opelopabga ⊢ A ∈ C ∧ B ∈ D → A B ∈ x y | x ∈ C ∧ y ∈ D ∧ φ ↔ A ∈ C ∧ B ∈ D ∧ ψ
7 6 bianabs ⊢ A ∈ C ∧ B ∈ D → A B ∈ x y | x ∈ C ∧ y ∈ D ∧ φ ↔ ψ