Metamath Proof Explorer


Theorem elab4g

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 17-Oct-2012)

Ref Expression
Hypotheses elab4g.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
elab4g.2 𝐵 = { 𝑥𝜑 }
Assertion elab4g ( 𝐴𝐵 ↔ ( 𝐴 ∈ V ∧ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elab4g.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
2 elab4g.2 𝐵 = { 𝑥𝜑 }
3 elex ( 𝐴𝐵𝐴 ∈ V )
4 1 2 elab2g ( 𝐴 ∈ V → ( 𝐴𝐵𝜓 ) )
5 3 4 biadanii ( 𝐴𝐵 ↔ ( 𝐴 ∈ V ∧ 𝜓 ) )