Metamath Proof Explorer


Theorem bj-epelg

Description: The membership relation and the membership predicate agree when the "containing" class is a set. General version of epel and closed form of epeli . (Contributed by Scott Fenton, 27-Mar-2011) (Revised by Mario Carneiro, 28-Apr-2015) TODO: move it to the main section after reordering to have brrelex1i available. (Proof shortened by BJ, 14-Jul-2023) (Proof modification is discouraged.)

Ref Expression
Assertion bj-epelg
|- ( B e. V -> ( A _E B <-> A e. B ) )

Proof

Step Hyp Ref Expression
1 rele
 |-  Rel _E
2 1 brrelex1i
 |-  ( A _E B -> A e. _V )
3 2 a1i
 |-  ( B e. V -> ( A _E B -> A e. _V ) )
4 elex
 |-  ( A e. B -> A e. _V )
5 4 a1i
 |-  ( B e. V -> ( A e. B -> A e. _V ) )
6 eleq12
 |-  ( ( x = A /\ y = B ) -> ( x e. y <-> A e. B ) )
7 df-eprel
 |-  _E = { <. x , y >. | x e. y }
8 6 7 brabga
 |-  ( ( A e. _V /\ B e. V ) -> ( A _E B <-> A e. B ) )
9 8 expcom
 |-  ( B e. V -> ( A e. _V -> ( A _E B <-> A e. B ) ) )
10 3 5 9 pm5.21ndd
 |-  ( B e. V -> ( A _E B <-> A e. B ) )