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) Remove dependency on ax-13 . (Revised by Steven Nguyen, 23-Nov-2022)

Ref Expression
Hypothesis elabg.1 x = A φ ψ
Assertion elabg A V A x | φ ψ

Proof

Step Hyp Ref Expression
1 elabg.1 x = A φ ψ
2 nfab1 _ x x | φ
3 2 nfel2 x A x | φ
4 nfv x ψ
5 3 4 nfbi x A x | φ ψ
6 eleq1 x = A x x | φ A x | φ
7 6 1 bibi12d x = A x x | φ φ A x | φ ψ
8 abid x x | φ φ
9 5 7 8 vtoclg1f A V A x | φ ψ