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 → ( 𝜑𝜓 ) )
Assertion elabg ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elabg.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
2 1 ax-gen 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
3 elabgt ( ( 𝐴𝑉 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )
4 2 3 mpan2 ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )