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