Metamath Proof Explorer


Theorem bj-inrab

Description: Generalization of inrab . (Contributed by BJ, 21-Apr-2019)

Ref Expression
Assertion bj-inrab ⊢ x ∈ A | φ ∩ x ∈ B | ψ = x ∈ A ∩ B | φ ∧ ψ

Proof

Step Hyp Ref Expression
1 an4 ⊢ x ∈ A ∧ φ ∧ x ∈ B ∧ ψ ↔ x ∈ A ∧ x ∈ B ∧ φ ∧ ψ
2 elin ⊢ x ∈ A ∩ B ↔ x ∈ A ∧ x ∈ B
3 2 anbi1i ⊢ x ∈ A ∩ B ∧ φ ∧ ψ ↔ x ∈ A ∧ x ∈ B ∧ φ ∧ ψ
4 1 3 bitr4i ⊢ x ∈ A ∧ φ ∧ x ∈ B ∧ ψ ↔ x ∈ A ∩ B ∧ φ ∧ ψ
5 4 abbii ⊢ x | x ∈ A ∧ φ ∧ x ∈ B ∧ ψ = x | x ∈ A ∩ B ∧ φ ∧ ψ
6 df-rab ⊢ x ∈ A | φ = x | x ∈ A ∧ φ
7 df-rab ⊢ x ∈ B | ψ = x | x ∈ B ∧ ψ
8 6 7 ineq12i ⊢ x ∈ A | φ ∩ x ∈ B | ψ = x | x ∈ A ∧ φ ∩ x | x ∈ B ∧ ψ
9 inab ⊢ x | x ∈ A ∧ φ ∩ x | x ∈ B ∧ ψ = x | x ∈ A ∧ φ ∧ x ∈ B ∧ ψ
10 8 9 eqtri ⊢ x ∈ A | φ ∩ x ∈ B | ψ = x | x ∈ A ∧ φ ∧ x ∈ B ∧ ψ
11 df-rab ⊢ x ∈ A ∩ B | φ ∧ ψ = x | x ∈ A ∩ B ∧ φ ∧ ψ
12 5 10 11 3eqtr4i ⊢ x ∈ A | φ ∩ x ∈ B | ψ = x ∈ A ∩ B | φ ∧ ψ