Metamath Proof Explorer


Theorem opabssi

Description: Sufficient condition for a collection of ordered pairs to be a subclass of a relation. (Contributed by Peter Mazsa, 21-Oct-2019) (Revised by Thierry Arnoux, 18-Feb-2022)

Ref Expression
Hypothesis opabssi.1
|- ( ph -> <. x , y >. e. A )
Assertion opabssi
|- { <. x , y >. | ph } C_ A

Proof

Step Hyp Ref Expression
1 opabssi.1
 |-  ( ph -> <. x , y >. e. A )
2 df-opab
 |-  { <. x , y >. | ph } = { z | E. x E. y ( z = <. x , y >. /\ ph ) }
3 eleq1
 |-  ( z = <. x , y >. -> ( z e. A <-> <. x , y >. e. A ) )
4 3 biimprd
 |-  ( z = <. x , y >. -> ( <. x , y >. e. A -> z e. A ) )
5 4 1 impel
 |-  ( ( z = <. x , y >. /\ ph ) -> z e. A )
6 5 exlimivv
 |-  ( E. x E. y ( z = <. x , y >. /\ ph ) -> z e. A )
7 6 abssi
 |-  { z | E. x E. y ( z = <. x , y >. /\ ph ) } C_ A
8 2 7 eqsstri
 |-  { <. x , y >. | ph } C_ A