Metamath Proof Explorer


Theorem elab2g

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 13-Sep-1995)

Ref Expression
Hypotheses elab2g.1
|- ( x = A -> ( ph <-> ps ) )
elab2g.2
|- B = { x | ph }
Assertion elab2g
|- ( A e. V -> ( A e. B <-> ps ) )

Proof

Step Hyp Ref Expression
1 elab2g.1
 |-  ( x = A -> ( ph <-> ps ) )
2 elab2g.2
 |-  B = { x | ph }
3 2 eleq2i
 |-  ( A e. B <-> A e. { x | ph } )
4 1 elabg
 |-  ( A e. V -> ( A e. { x | ph } <-> ps ) )
5 3 4 syl5bb
 |-  ( A e. V -> ( A e. B <-> ps ) )