Metamath Proof Explorer


Theorem eliminable-abelab

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): abstraction belongs to abstraction. (Contributed by BJ, 30-Apr-2024) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion eliminable-abelab ⊢ x | φ ∈ y | ψ ↔ ∃ z ∀ t t ∈ z ↔ t x φ ∧ z y ψ

Proof

Step Hyp Ref Expression
1 dfclel ⊢ x | φ ∈ y | ψ ↔ ∃ z z = x | φ ∧ z ∈ y | ψ
2 eliminable-veqab ⊢ z = x | φ ↔ ∀ t t ∈ z ↔ t x φ
3 eliminable-velab ⊢ z ∈ y | ψ ↔ z y ψ
4 2 3 anbi12i ⊢ z = x | φ ∧ z ∈ y | ψ ↔ ∀ t t ∈ z ↔ t x φ ∧ z y ψ
5 4 exbii ⊢ ∃ z z = x | φ ∧ z ∈ y | ψ ↔ ∃ z ∀ t t ∈ z ↔ t x φ ∧ z y ψ
6 1 5 bitri ⊢ x | φ ∈ y | ψ ↔ ∃ z ∀ t t ∈ z ↔ t x φ ∧ z y ψ