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 1 ax-gen x x = A φ ψ
3 elabgt A V x x = A φ ψ A x | φ ψ
4 2 3 mpan2 A V A x | φ ψ