Metamath Proof Explorer


Theorem elabgf

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. This version has bound-variable hypotheses in place of distinct variable restrictions. (Contributed by NM, 21-Sep-2003) (Revised by Mario Carneiro, 12-Oct-2016)

Ref Expression
Hypotheses elabgf.1
|- F/_ x A
elabgf.2
|- F/ x ps
elabgf.3
|- ( x = A -> ( ph <-> ps ) )
Assertion elabgf
|- ( A e. B -> ( A e. { x | ph } <-> ps ) )

Proof

Step Hyp Ref Expression
1 elabgf.1
 |-  F/_ x A
2 elabgf.2
 |-  F/ x ps
3 elabgf.3
 |-  ( x = A -> ( ph <-> ps ) )
4 nfab1
 |-  F/_ x { x | ph }
5 1 4 nfel
 |-  F/ x A e. { x | ph }
6 5 2 nfbi
 |-  F/ x ( A e. { x | ph } <-> ps )
7 eleq1
 |-  ( x = A -> ( x e. { x | ph } <-> A e. { x | ph } ) )
8 7 3 bibi12d
 |-  ( x = A -> ( ( x e. { x | ph } <-> ph ) <-> ( A e. { x | ph } <-> ps ) ) )
9 abid
 |-  ( x e. { x | ph } <-> ph )
10 1 6 8 9 vtoclgf
 |-  ( A e. B -> ( A e. { x | ph } <-> ps ) )