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Metamath Proof Explorer

Theorem elabgf

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. This version has bound-variable hypotheses in place of distinct variable restrictions. (Contributed by NM, 21-Sep-2003) (Revised by Mario Carneiro, 12-Oct-2016)

Ref Expression
Hypotheses elabgf.1 ⊢ Ⅎ _ x A
elabgf.2 ⊢ Ⅎ x ψ
elabgf.3 ⊢ x = A → φ ↔ ψ
Assertion elabgf ⊢ A ∈ B → A ∈ x | φ ↔ ψ

Proof

Step Hyp Ref Expression
1 elabgf.1 ⊢ Ⅎ _ x A
2 elabgf.2 ⊢ Ⅎ x ψ
3 elabgf.3 ⊢ x = A → φ ↔ ψ
4 nfab1 ⊢ Ⅎ _ x x | φ
5 1 4 nfel ⊢ Ⅎ x A ∈ x | φ
6 5 2 nfbi ⊢ Ⅎ x A ∈ x | φ ↔ ψ
7 eleq1 ⊢ x = A → x ∈ x | φ ↔ A ∈ x | φ
8 7 3 bibi12d ⊢ x = A → x ∈ x | φ ↔ φ ↔ A ∈ x | φ ↔ ψ
9 abid ⊢ x ∈ x | φ ↔ φ
10 1 6 8 9 vtoclgf ⊢ A ∈ B → A ∈ x | φ ↔ ψ