Metamath Proof Explorer


Theorem rankxplim2

Description: If the rank of a Cartesian product is a limit ordinal, so is the rank of the union of its arguments. (Contributed by NM, 19-Sep-2006)

Ref Expression
Hypotheses rankxplim.1 ⊢ A ∈ V
rankxplim.2 ⊢ B ∈ V
Assertion rankxplim2 ⊢ Lim ⁡ rank ⁡ A × B → Lim ⁡ rank ⁡ A ∪ B

Proof

Step Hyp Ref Expression
1 rankxplim.1 ⊢ A ∈ V
2 rankxplim.2 ⊢ B ∈ V
3 0ellim ⊢ Lim ⁡ rank ⁡ A × B → ∅ ∈ rank ⁡ A × B
4 n0i ⊢ ∅ ∈ rank ⁡ A × B → ¬ rank ⁡ A × B = ∅
5 3 4 syl ⊢ Lim ⁡ rank ⁡ A × B → ¬ rank ⁡ A × B = ∅
6 df-ne ⊢ A × B ≠ ∅ ↔ ¬ A × B = ∅
7 1 2 xpex ⊢ A × B ∈ V
8 7 rankeq0 ⊢ A × B = ∅ ↔ rank ⁡ A × B = ∅
9 8 notbii ⊢ ¬ A × B = ∅ ↔ ¬ rank ⁡ A × B = ∅
10 6 9 bitr2i ⊢ ¬ rank ⁡ A × B = ∅ ↔ A × B ≠ ∅
11 5 10 sylib ⊢ Lim ⁡ rank ⁡ A × B → A × B ≠ ∅
12 limuni2 ⊢ Lim ⁡ rank ⁡ A × B → Lim ⁡ ⋃ rank ⁡ A × B
13 limuni2 ⊢ Lim ⁡ ⋃ rank ⁡ A × B → Lim ⁡ ⋃ ⋃ rank ⁡ A × B
14 12 13 syl ⊢ Lim ⁡ rank ⁡ A × B → Lim ⁡ ⋃ ⋃ rank ⁡ A × B
15 rankuni ⊢ rank ⁡ ⋃ ⋃ A × B = ⋃ rank ⁡ ⋃ A × B
16 rankuni ⊢ rank ⁡ ⋃ A × B = ⋃ rank ⁡ A × B
17 16 unieqi ⊢ ⋃ rank ⁡ ⋃ A × B = ⋃ ⋃ rank ⁡ A × B
18 15 17 eqtr2i ⊢ ⋃ ⋃ rank ⁡ A × B = rank ⁡ ⋃ ⋃ A × B
19 unixp ⊢ A × B ≠ ∅ → ⋃ ⋃ A × B = A ∪ B
20 19 fveq2d ⊢ A × B ≠ ∅ → rank ⁡ ⋃ ⋃ A × B = rank ⁡ A ∪ B
21 18 20 eqtrid ⊢ A × B ≠ ∅ → ⋃ ⋃ rank ⁡ A × B = rank ⁡ A ∪ B
22 limeq ⊢ ⋃ ⋃ rank ⁡ A × B = rank ⁡ A ∪ B → Lim ⁡ ⋃ ⋃ rank ⁡ A × B ↔ Lim ⁡ rank ⁡ A ∪ B
23 21 22 syl ⊢ A × B ≠ ∅ → Lim ⁡ ⋃ ⋃ rank ⁡ A × B ↔ Lim ⁡ rank ⁡ A ∪ B
24 14 23 imbitrid ⊢ A × B ≠ ∅ → Lim ⁡ rank ⁡ A × B → Lim ⁡ rank ⁡ A ∪ B
25 11 24 mpcom ⊢ Lim ⁡ rank ⁡ A × B → Lim ⁡ rank ⁡ A ∪ B