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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 GG, 12-Oct-2024) (Proof shortened by Wolf Lammen, 11-May-2025) (Proof shortened by SN, 1-Dec-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 pm5.74 ⊢ x = A → φ ↔ ψ ↔ x = A → φ ↔ x = A → ψ
3 2 biimpi ⊢ x = A → φ ↔ ψ → x = A → φ ↔ x = A → ψ
4 3 alimi ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ x = A → ψ
5 albi ⊢ ∀ x x = A → φ ↔ x = A → ψ → ∀ x x = A → φ ↔ ∀ x x = A → ψ
6 4 5 syl ⊢ ∀ x x = A → φ ↔ ψ → ∀ x x = A → φ ↔ ∀ x x = A → ψ
7 1 6 sylan9bb ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ∀ x x = A → ψ
8 19.23v ⊢ ∀ x x = A → ψ ↔ ∃ x x = A → ψ
9 elisset ⊢ A ∈ B → ∃ x x = A
10 pm5.5 ⊢ ∃ x x = A → ∃ x x = A → ψ ↔ ψ
11 9 10 syl ⊢ A ∈ B → ∃ x x = A → ψ ↔ ψ
12 8 11 bitrid ⊢ A ∈ B → ∀ x x = A → ψ ↔ ψ
13 12 adantr ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → ∀ x x = A → ψ ↔ ψ
14 7 13 bitrd ⊢ A ∈ B ∧ ∀ x x = A → φ ↔ ψ → A ∈ x | φ ↔ ψ