Metamath Proof Explorer


Theorem elabg

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. (Contributed by NM, 14-Apr-1995) Avoid ax-13 . (Revised by SN, 23-Nov-2022) Avoid ax-10 , ax-11 , ax-12 . (Revised by SN, 5-Oct-2024)

Ref Expression
Hypothesis elabg.1
|- ( x = A -> ( ph <-> ps ) )
Assertion elabg
|- ( A e. V -> ( A e. { x | ph } <-> ps ) )

Proof

Step Hyp Ref Expression
1 elabg.1
 |-  ( x = A -> ( ph <-> ps ) )
2 elab6g
 |-  ( A e. V -> ( A e. { x | ph } <-> A. x ( x = A -> ph ) ) )
3 1 pm5.74i
 |-  ( ( x = A -> ph ) <-> ( x = A -> ps ) )
4 3 albii
 |-  ( A. x ( x = A -> ph ) <-> A. x ( x = A -> ps ) )
5 19.23v
 |-  ( A. x ( x = A -> ps ) <-> ( E. x x = A -> ps ) )
6 4 5 bitri
 |-  ( A. x ( x = A -> ph ) <-> ( E. x x = A -> ps ) )
7 elisset
 |-  ( A e. V -> E. x x = A )
8 pm5.5
 |-  ( E. x x = A -> ( ( E. x x = A -> ps ) <-> ps ) )
9 7 8 syl
 |-  ( A e. V -> ( ( E. x x = A -> ps ) <-> ps ) )
10 6 9 syl5bb
 |-  ( A e. V -> ( A. x ( x = A -> ph ) <-> ps ) )
11 2 10 bitrd
 |-  ( A e. V -> ( A e. { x | ph } <-> ps ) )