Metamath Proof Explorer


Theorem elab2

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

Ref Expression
Hypotheses elab2.1
|- A e. _V
elab2.2
|- ( x = A -> ( ph <-> ps ) )
elab2.3
|- B = { x | ph }
Assertion elab2
|- ( A e. B <-> ps )

Proof

Step Hyp Ref Expression
1 elab2.1
 |-  A e. _V
2 elab2.2
 |-  ( x = A -> ( ph <-> ps ) )
3 elab2.3
 |-  B = { x | ph }
4 2 3 elab2g
 |-  ( A e. _V -> ( A e. B <-> ps ) )
5 1 4 ax-mp
 |-  ( A e. B <-> ps )