Metamath Proof Explorer


Theorem alexeqg

Description: Two ways to express substitution of A for x in ph . This is the analogue for classes of sbalex . (Contributed by NM, 2-Mar-1995) (Revised by BJ, 27-Apr-2019)

Ref Expression
Assertion alexeqg ⊢ A ∈ V → ∀ x x = A → φ ↔ ∃ x x = A ∧ φ

Proof

Step Hyp Ref Expression
1 eqeq2 ⊢ y = A → x = y ↔ x = A
2 1 anbi1d ⊢ y = A → x = y ∧ φ ↔ x = A ∧ φ
3 2 exbidv ⊢ y = A → ∃ x x = y ∧ φ ↔ ∃ x x = A ∧ φ
4 1 imbi1d ⊢ y = A → x = y → φ ↔ x = A → φ
5 4 albidv ⊢ y = A → ∀ x x = y → φ ↔ ∀ x x = A → φ
6 sbalex ⊢ ∃ x x = y ∧ φ ↔ ∀ x x = y → φ
7 3 5 6 vtoclbg ⊢ A ∈ V → ∃ x x = A ∧ φ ↔ ∀ x x = A → φ
8 7 bicomd ⊢ A ∈ V → ∀ x x = A → φ ↔ ∃ x x = A ∧ φ