Metamath Proof Explorer


Theorem 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) (Proof shortened by BJ, 14-Jul-2023)

Ref Expression
Assertion epelg B V A E B A B

Proof

Step Hyp Ref Expression
1 df-br A E B A B E
2 0nelopab ¬ x y | x y
3 df-eprel E = x y | x y
4 3 eqcomi x y | x y = E
5 4 eleq2i x y | x y E
6 2 5 mtbi ¬ E
7 eleq1 A B = A B E E
8 6 7 mtbiri A B = ¬ A B E
9 8 con2i A B E ¬ A B =
10 opprc1 ¬ A V A B =
11 9 10 nsyl2 A B E A V
12 1 11 sylbi A E B A V
13 12 a1i B V A E B A V
14 elex A B A V
15 14 a1i B V A B A V
16 eleq12 x = A y = B x y A B
17 16 3 brabga A V B V A E B A B
18 17 expcom B V A V A E B A B
19 13 15 18 pm5.21ndd B V A E B A B