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 φ ψ
Assertion elabg A V A x | φ ψ

Proof

Step Hyp Ref Expression
1 elabg.1 x = A φ ψ
2 elab6g A V A x | φ x x = A φ
3 1 pm5.74i x = A φ x = A ψ
4 3 albii x x = A φ x x = A ψ
5 19.23v x x = A ψ x x = A ψ
6 4 5 bitri x x = A φ x x = A ψ
7 elisset A V x x = A
8 pm5.5 x x = A x x = A ψ ψ
9 7 8 syl A V x x = A ψ ψ
10 6 9 syl5bb A V x x = A φ ψ
11 2 10 bitrd A V A x | φ ψ