Metamath Proof Explorer

Theorem elabgt

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg .) (Contributed by NM, 7-Nov-2005) (Proof shortened by Andrew Salmon, 8-Jun-2011) Reduce axiom usage. (Revised by Gino Giotto, 12-Oct-2024)

Ref Expression
Assertion elabgt ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ψ

Proof

Step Hyp Ref Expression
1 elab6g ⊢ A ∈ B → A ∈ x | φ ↔ ∀ x x = A → φ
2 1 adantr ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ∀ x x = A → φ
3 elisset ⊢ A ∈ B → ∃ x x = A
4 biimp ⊢ φ ↔ ψ → φ → ψ
5 4 imim3i ⊢ x = A → φ ↔ ψ → x = A → φ → x = A → ψ
6 5 al2imi ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ∀ x x = A → ψ
7 19.23v ⊢ ∀ x x = A → ψ ↔ ∃ x x = A → ψ
8 6 7 syl6ib ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ∃ x x = A → ψ
9 8 com3r ⊢ ∃ x x = A → ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ψ
10 biimpr ⊢ φ ↔ ψ → ψ → φ
11 10 imim2i ⊢ x = A → φ ↔ ψ → x = A → ψ → φ
12 11 alimi ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → ψ → φ
13 bi2.04 ⊢ x = A → ψ → φ ↔ ψ → x = A → φ
14 13 albii ⊢ ∀ x x = A → ψ → φ ↔ ∀ x ψ → x = A → φ
15 19.21v ⊢ ∀ x ψ → x = A → φ ↔ ψ → ∀ x x = A → φ
16 14 15 sylbb ⊢ ∀ x x = A → ψ → φ → ψ → ∀ x x = A → φ
17 12 16 syl ⊢ ∀ x x = A → φ ↔ ψ → ψ → ∀ x x = A → φ
18 17 a1i ⊢ ∃ x x = A → ∀ x x = A → φ ↔ ψ → ψ → ∀ x x = A → φ
19 9 18 impbidd ⊢ ∃ x x = A → ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ ψ
20 3 19 syl ⊢ A ∈ B → ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ ψ
21 20 imp ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ ψ
22 2 21 bitrd ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ψ