Metamath Proof Explorer


Theorem elab6g

Description: Membership in a class abstraction. Class version of sb6 . (Contributed by SN, 5-Oct-2024)

Ref Expression
Assertion elab6g ⊢ A ∈ V → A ∈ x | φ ↔ ∀ x x = A → φ

Proof

Step Hyp Ref Expression
1 eleq1 ⊢ y = A → y ∈ x | φ ↔ A ∈ x | φ
2 eqeq2 ⊢ y = A → x = y ↔ x = A
3 2 imbi1d ⊢ y = A → x = y → φ ↔ x = A → φ
4 3 albidv ⊢ y = A → ∀ x x = y → φ ↔ ∀ x x = A → φ
5 df-clab ⊢ y ∈ x | φ ↔ y x φ
6 sb6 ⊢ y x φ ↔ ∀ x x = y → φ
7 5 6 bitri ⊢ y ∈ x | φ ↔ ∀ x x = y → φ
8 1 4 7 vtoclbg ⊢ A ∈ V → A ∈ x | φ ↔ ∀ x x = A → φ