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 e. On <-> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )

Proof

Step Hyp Ref Expression
1 elex 
 |-  ( A e. On -> A e. _V )
2 eloni 
 |-  ( A e. On -> Ord A )
3 ordzsl 
 |-  ( Ord A <-> ( A = (/) \/ E. x e. On A = suc x \/ Lim A ) )
4 3mix1 
 |-  ( A = (/) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
5 4 adantl 
 |-  ( ( A e. _V /\ A = (/) ) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
6 3mix2 
 |-  ( E. x e. On A = suc x -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
7 6 adantl 
 |-  ( ( A e. _V /\ E. x e. On A = suc x ) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
8 3mix3 
 |-  ( ( A e. _V /\ Lim A ) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
9 5 7 8 3jaodan 
 |-  ( ( A e. _V /\ ( A = (/) \/ E. x e. On A = suc x \/ Lim A ) ) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
10 3 9 sylan2b 
 |-  ( ( A e. _V /\ Ord A ) -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
11 1 2 10 syl2anc 
 |-  ( A e. On -> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )
12 0elon 
 |-  (/) e. On
13 eleq1 
 |-  ( A = (/) -> ( A e. On <-> (/) e. On ) )
14 12 13 mpbiri 
 |-  ( A = (/) -> A e. On )
15 onsuc 
 |-  ( x e. On -> suc x e. On )
16 eleq1 
 |-  ( A = suc x -> ( A e. On <-> suc x e. On ) )
17 15 16 syl5ibrcom 
 |-  ( x e. On -> ( A = suc x -> A e. On ) )
18 17 rexlimiv 
 |-  ( E. x e. On A = suc x -> A e. On )
19 limelon 
 |-  ( ( A e. _V /\ Lim A ) -> A e. On )
20 14 18 19 3jaoi 
 |-  ( ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) -> A e. On )
21 11 20 impbii 
 |-  ( A e. On <-> ( A = (/) \/ E. x e. On A = suc x \/ ( A e. _V /\ Lim A ) ) )