Metamath Proof Explorer


Theorem iununi

Description: A relationship involving union and indexed union. Exercise 25 of Enderton p. 33. (Contributed by NM, 25-Nov-2003) (Proof shortened by Mario Carneiro, 17-Nov-2016)

Ref Expression
Assertion iununi ⊢ B = ∅ → A = ∅ ↔ A ∪ ⋃ B = ⋃ x ∈ B A ∪ x

Proof

Step Hyp Ref Expression
1 df-ne ⊢ B ≠ ∅ ↔ ¬ B = ∅
2 iunconst ⊢ B ≠ ∅ → ⋃ x ∈ B A = A
3 1 2 sylbir ⊢ ¬ B = ∅ → ⋃ x ∈ B A = A
4 iun0 ⊢ ⋃ x ∈ B ∅ = ∅
5 id ⊢ A = ∅ → A = ∅
6 5 iuneq2d ⊢ A = ∅ → ⋃ x ∈ B A = ⋃ x ∈ B ∅
7 4 6 5 3eqtr4a ⊢ A = ∅ → ⋃ x ∈ B A = A
8 3 7 ja ⊢ B = ∅ → A = ∅ → ⋃ x ∈ B A = A
9 8 eqcomd ⊢ B = ∅ → A = ∅ → A = ⋃ x ∈ B A
10 9 uneq1d ⊢ B = ∅ → A = ∅ → A ∪ ⋃ x ∈ B x = ⋃ x ∈ B A ∪ ⋃ x ∈ B x
11 uniiun ⊢ ⋃ B = ⋃ x ∈ B x
12 11 uneq2i ⊢ A ∪ ⋃ B = A ∪ ⋃ x ∈ B x
13 iunun ⊢ ⋃ x ∈ B A ∪ x = ⋃ x ∈ B A ∪ ⋃ x ∈ B x
14 10 12 13 3eqtr4g ⊢ B = ∅ → A = ∅ → A ∪ ⋃ B = ⋃ x ∈ B A ∪ x
15 unieq ⊢ B = ∅ → ⋃ B = ⋃ ∅
16 uni0 ⊢ ⋃ ∅ = ∅
17 15 16 eqtrdi ⊢ B = ∅ → ⋃ B = ∅
18 17 uneq2d ⊢ B = ∅ → A ∪ ⋃ B = A ∪ ∅
19 un0 ⊢ A ∪ ∅ = A
20 18 19 eqtrdi ⊢ B = ∅ → A ∪ ⋃ B = A
21 iuneq1 ⊢ B = ∅ → ⋃ x ∈ B A ∪ x = ⋃ x ∈ ∅ A ∪ x
22 0iun ⊢ ⋃ x ∈ ∅ A ∪ x = ∅
23 21 22 eqtrdi ⊢ B = ∅ → ⋃ x ∈ B A ∪ x = ∅
24 20 23 eqeq12d ⊢ B = ∅ → A ∪ ⋃ B = ⋃ x ∈ B A ∪ x ↔ A = ∅
25 24 biimpcd ⊢ A ∪ ⋃ B = ⋃ x ∈ B A ∪ x → B = ∅ → A = ∅
26 14 25 impbii ⊢ B = ∅ → A = ∅ ↔ A ∪ ⋃ B = ⋃ x ∈ B A ∪ x