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) (Proof shortend by Wolf Lammen, 11-May-2025.)

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 imbitrdi ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ∃ x x = A → ψ
9 3 8 syl7 ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → A ∈ B → ψ
10 9 com3r ⊢ A ∈ B → ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ψ
11 10 imp ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ → ψ
12 biimpr ⊢ φ ↔ ψ → ψ → φ
13 12 imim2i ⊢ x = A → φ ↔ ψ → x = A → ψ → φ
14 13 com23 ⊢ x = A → φ ↔ ψ → ψ → x = A → φ
15 14 alimi ⊢ ∀ x x = A → φ ↔ ψ → ∀ x ψ → x = A → φ
16 19.21v ⊢ ∀ x ψ → x = A → φ ↔ ψ → ∀ x x = A → φ
17 15 16 sylib ⊢ ∀ x x = A → φ ↔ ψ → ψ → ∀ x x = A → φ
18 17 adantl ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → ψ → ∀ x x = A → φ
19 11 18 impbid ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ ψ
20 2 19 bitrd ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ψ