Metamath Proof Explorer


Theorem rankeq0b

Description: A set is empty iff its rank is empty. (Contributed by Mario Carneiro, 17-Nov-2014)

Ref Expression
Assertion rankeq0b ⊢ A ∈ ⋃ R1 On → A = ∅ ↔ rank ⁡ A = ∅

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ A = ∅ → rank ⁡ A = rank ⁡ ∅
2 r1funlim ⊢ Fun ⁡ R1 ∧ Lim ⁡ dom ⁡ R1
3 2 simpri ⊢ Lim ⁡ dom ⁡ R1
4 limomss ⊢ Lim ⁡ dom ⁡ R1 → ω ⊆ dom ⁡ R1
5 3 4 ax-mp ⊢ ω ⊆ dom ⁡ R1
6 peano1 ⊢ ∅ ∈ ω
7 5 6 sselii ⊢ ∅ ∈ dom ⁡ R1
8 rankonid ⊢ ∅ ∈ dom ⁡ R1 ↔ rank ⁡ ∅ = ∅
9 7 8 mpbi ⊢ rank ⁡ ∅ = ∅
10 1 9 eqtrdi ⊢ A = ∅ → rank ⁡ A = ∅
11 eqimss ⊢ rank ⁡ A = ∅ → rank ⁡ A ⊆ ∅
12 11 adantl ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → rank ⁡ A ⊆ ∅
13 simpl ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → A ∈ ⋃ R1 On
14 rankr1bg ⊢ A ∈ ⋃ R1 On ∧ ∅ ∈ dom ⁡ R1 → A ⊆ R1 ⁡ ∅ ↔ rank ⁡ A ⊆ ∅
15 13 7 14 sylancl ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → A ⊆ R1 ⁡ ∅ ↔ rank ⁡ A ⊆ ∅
16 12 15 mpbird ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → A ⊆ R1 ⁡ ∅
17 r10 ⊢ R1 ⁡ ∅ = ∅
18 16 17 sseqtrdi ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → A ⊆ ∅
19 ss0 ⊢ A ⊆ ∅ → A = ∅
20 18 19 syl ⊢ A ∈ ⋃ R1 On ∧ rank ⁡ A = ∅ → A = ∅
21 20 ex ⊢ A ∈ ⋃ R1 On → rank ⁡ A = ∅ → A = ∅
22 10 21 impbid2 ⊢ A ∈ ⋃ R1 On → A = ∅ ↔ rank ⁡ A = ∅