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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 ) -> ( ph <-> ps ) )
Assertion opelopab2a 
|- ( ( A e. C /\ B e. D ) -> ( <. A , B >. e. { <. x , y >. | ( ( x e. C /\ y e. D ) /\ ph ) } <-> ps ) )

Proof

Step Hyp Ref Expression
1 opelopabga.1 
 |-  ( ( x = A /\ y = B ) -> ( ph <-> ps ) )
2 eleq1 
 |-  ( x = A -> ( x e. C <-> A e. C ) )
3 eleq1 
 |-  ( y = B -> ( y e. D <-> B e. D ) )
4 2 3 bi2anan9 
 |-  ( ( x = A /\ y = B ) -> ( ( x e. C /\ y e. D ) <-> ( A e. C /\ B e. D ) ) )
5 4 1 anbi12d 
 |-  ( ( x = A /\ y = B ) -> ( ( ( x e. C /\ y e. D ) /\ ph ) <-> ( ( A e. C /\ B e. D ) /\ ps ) ) )
6 5 opelopabga 
 |-  ( ( A e. C /\ B e. D ) -> ( <. A , B >. e. { <. x , y >. | ( ( x e. C /\ y e. D ) /\ ph ) } <-> ( ( A e. C /\ B e. D ) /\ ps ) ) )
7 6 bianabs 
 |-  ( ( A e. C /\ B e. D ) -> ( <. A , B >. e. { <. x , y >. | ( ( x e. C /\ y e. D ) /\ ph ) } <-> ps ) )