Metamath Proof Explorer


Theorem disjrdx

Description: Re-index a disjunct collection statement. (Contributed by Thierry Arnoux, 7-Apr-2017)

Ref Expression
Hypotheses disjrdx.1 ⊢ φ → F : A ⟶ 1-1 onto C
disjrdx.2 ⊢ φ ∧ y = F ⁡ x → D = B
Assertion disjrdx ⊢ φ → Disj x ∈ A B ↔ Disj y ∈ C D

Proof

Step Hyp Ref Expression
1 disjrdx.1 ⊢ φ → F : A ⟶ 1-1 onto C
2 disjrdx.2 ⊢ φ ∧ y = F ⁡ x → D = B
3 f1of ⊢ F : A ⟶ 1-1 onto C → F : A ⟶ C
4 1 3 syl ⊢ φ → F : A ⟶ C
5 4 ffvelcdmda ⊢ φ ∧ x ∈ A → F ⁡ x ∈ C
6 f1ofveu ⊢ F : A ⟶ 1-1 onto C ∧ y ∈ C → ∃! x ∈ A F ⁡ x = y
7 1 6 sylan ⊢ φ ∧ y ∈ C → ∃! x ∈ A F ⁡ x = y
8 eqcom ⊢ F ⁡ x = y ↔ y = F ⁡ x
9 8 reubii ⊢ ∃! x ∈ A F ⁡ x = y ↔ ∃! x ∈ A y = F ⁡ x
10 7 9 sylib ⊢ φ ∧ y ∈ C → ∃! x ∈ A y = F ⁡ x
11 2 eleq2d ⊢ φ ∧ y = F ⁡ x → z ∈ D ↔ z ∈ B
12 5 10 11 rmoxfrd ⊢ φ → ∃* y ∈ C z ∈ D ↔ ∃* x ∈ A z ∈ B
13 12 bicomd ⊢ φ → ∃* x ∈ A z ∈ B ↔ ∃* y ∈ C z ∈ D
14 13 albidv ⊢ φ → ∀ z ∃* x ∈ A z ∈ B ↔ ∀ z ∃* y ∈ C z ∈ D
15 df-disj ⊢ Disj x ∈ A B ↔ ∀ z ∃* x ∈ A z ∈ B
16 df-disj ⊢ Disj y ∈ C D ↔ ∀ z ∃* y ∈ C z ∈ D
17 14 15 16 3bitr4g ⊢ φ → Disj x ∈ A B ↔ Disj y ∈ C D