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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 → ( 𝜑 ↔ 𝜓 ) )
elab4g.2 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
Assertion elab4g ( 𝐴 ∈ 𝐵 ↔ ( 𝐴 ∈ V ∧ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elab4g.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 elab4g.2 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
3 elex ⊢ ( 𝐴 ∈ 𝐵 → 𝐴 ∈ V )
4 1 2 elab2g ⊢ ( 𝐴 ∈ V → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) )
5 3 4 biadanii ⊢ ( 𝐴 ∈ 𝐵 ↔ ( 𝐴 ∈ V ∧ 𝜓 ) )