Metamath Proof Explorer


Theorem elabf

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 1-Aug-1994) (Revised by Mario Carneiro, 12-Oct-2016)

Ref Expression
Hypotheses elabf.1 𝑥 𝜓
elabf.2 𝐴 ∈ V
elabf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion elabf ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 )

Proof

Step Hyp Ref Expression
1 elabf.1 𝑥 𝜓
2 elabf.2 𝐴 ∈ V
3 elabf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
4 nfcv 𝑥 𝐴
5 4 1 3 elabgf ( 𝐴 ∈ V → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )
6 2 5 ax-mp ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 )