Metamath Proof Explorer


Theorem opabssxp

Description: An abstraction relation is a subset of a related Cartesian product. (Contributed by NM, 16-Jul-1995)

Ref Expression
Assertion opabssxp
|- { <. x , y >. | ( ( x e. A /\ y e. B ) /\ ph ) } C_ ( A X. B )

Proof

Step Hyp Ref Expression
1 simpl
 |-  ( ( ( x e. A /\ y e. B ) /\ ph ) -> ( x e. A /\ y e. B ) )
2 1 ssopab2i
 |-  { <. x , y >. | ( ( x e. A /\ y e. B ) /\ ph ) } C_ { <. x , y >. | ( x e. A /\ y e. B ) }
3 df-xp
 |-  ( A X. B ) = { <. x , y >. | ( x e. A /\ y e. B ) }
4 2 3 sseqtrri
 |-  { <. x , y >. | ( ( x e. A /\ y e. B ) /\ ph ) } C_ ( A X. B )