Metamath Proof Explorer


Theorem resopab2

Description: Restriction of a class abstraction of ordered pairs. (Contributed by NM, 24-Aug-2007)

Ref Expression
Assertion resopab2 ⊢ A ⊆ B → x y | x ∈ B ∧ φ ↾ A = x y | x ∈ A ∧ φ

Proof

Step Hyp Ref Expression
1 resopab ⊢ x y | x ∈ B ∧ φ ↾ A = x y | x ∈ A ∧ x ∈ B ∧ φ
2 ssel ⊢ A ⊆ B → x ∈ A → x ∈ B
3 2 pm4.71d ⊢ A ⊆ B → x ∈ A ↔ x ∈ A ∧ x ∈ B
4 3 anbi1d ⊢ A ⊆ B → x ∈ A ∧ φ ↔ x ∈ A ∧ x ∈ B ∧ φ
5 anass ⊢ x ∈ A ∧ x ∈ B ∧ φ ↔ x ∈ A ∧ x ∈ B ∧ φ
6 4 5 bitr2di ⊢ A ⊆ B → x ∈ A ∧ x ∈ B ∧ φ ↔ x ∈ A ∧ φ
7 6 opabbidv ⊢ A ⊆ B → x y | x ∈ A ∧ x ∈ B ∧ φ = x y | x ∈ A ∧ φ
8 1 7 eqtrid ⊢ A ⊆ B → x y | x ∈ B ∧ φ ↾ A = x y | x ∈ A ∧ φ