Metamath Proof Explorer

Theorem onzsl

Description: An ordinal number is zero, a successor ordinal, or a limit ordinal number. (Contributed by NM, 1-Oct-2003) (Proof shortened by Andrew Salmon, 27-Aug-2011)

Ref Expression
Assertion onzsl ⊢A∈On↔A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A

Proof

Step Hyp Ref Expression
1 elex ⊢A∈On→A∈V
2 eloni ⊢A∈On→Ord⁡A
3 ordzsl ⊢Ord⁡A↔A=∅∨∃x∈OnA=suc⁡x∨Lim⁡A
4 3mix1 ⊢A=∅→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
5 4 adantl ⊢A∈V∧A=∅→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
6 3mix2 ⊢∃x∈OnA=suc⁡x→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
7 6 adantl ⊢A∈V∧∃x∈OnA=suc⁡x→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
8 3mix3 ⊢A∈V∧Lim⁡A→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
9 5 7 8 3jaodan ⊢A∈V∧A=∅∨∃x∈OnA=suc⁡x∨Lim⁡A→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
10 3 9 sylan2b ⊢A∈V∧Ord⁡A→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
11 1 2 10 syl2anc ⊢A∈On→A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A
12 0elon ⊢∅∈On
13 eleq1 ⊢A=∅→A∈On↔∅∈On
14 12 13 mpbiri ⊢A=∅→A∈On
15 onsuc ⊢x∈On→suc⁡x∈On
16 eleq1 ⊢A=suc⁡x→A∈On↔suc⁡x∈On
17 15 16 syl5ibrcom ⊢x∈On→A=suc⁡x→A∈On
18 17 rexlimiv ⊢∃x∈OnA=suc⁡x→A∈On
19 limelon ⊢A∈V∧Lim⁡A→A∈On
20 14 18 19 3jaoi ⊢A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A→A∈On
21 11 20 impbii ⊢A∈On↔A=∅∨∃x∈OnA=suc⁡x∨A∈V∧Lim⁡A