Metamath Proof Explorer


Theorem eltg3

Description: Membership in a topology generated by a basis. (Contributed by NM, 15-Jul-2006) (Proof shortened by Mario Carneiro, 30-Aug-2015)

Ref Expression
Assertion eltg3 B V A topGen B x x B A = x

Proof

Step Hyp Ref Expression
1 elfvdm A topGen B B dom topGen
2 inex1g B dom topGen B 𝒫 A V
3 1 2 syl A topGen B B 𝒫 A V
4 eltg4i A topGen B A = B 𝒫 A
5 inss1 B 𝒫 A B
6 sseq1 x = B 𝒫 A x B B 𝒫 A B
7 5 6 mpbiri x = B 𝒫 A x B
8 7 biantrurd x = B 𝒫 A A = x x B A = x
9 unieq x = B 𝒫 A x = B 𝒫 A
10 9 eqeq2d x = B 𝒫 A A = x A = B 𝒫 A
11 8 10 bitr3d x = B 𝒫 A x B A = x A = B 𝒫 A
12 3 4 11 spcedv A topGen B x x B A = x
13 eltg3i B V x B x topGen B
14 eleq1 A = x A topGen B x topGen B
15 13 14 syl5ibrcom B V x B A = x A topGen B
16 15 expimpd B V x B A = x A topGen B
17 16 exlimdv B V x x B A = x A topGen B
18 12 17 impbid2 B V A topGen B x x B A = x