Metamath Proof Explorer


Theorem df3o3

Description: Ordinal 3, fully expanded. (Contributed by RP, 8-Jul-2021)

Ref Expression
Assertion df3o3
|- 3o = { (/) , { (/) } , { (/) , { (/) } } }

Proof

Step Hyp Ref Expression
1 df-3o
 |-  3o = suc 2o
2 df2o2
 |-  2o = { (/) , { (/) } }
3 2 sneqi
 |-  { 2o } = { { (/) , { (/) } } }
4 2 3 uneq12i
 |-  ( 2o u. { 2o } ) = ( { (/) , { (/) } } u. { { (/) , { (/) } } } )
5 df-suc
 |-  suc 2o = ( 2o u. { 2o } )
6 df-tp
 |-  { (/) , { (/) } , { (/) , { (/) } } } = ( { (/) , { (/) } } u. { { (/) , { (/) } } } )
7 4 5 6 3eqtr4i
 |-  suc 2o = { (/) , { (/) } , { (/) , { (/) } } }
8 1 7 eqtri
 |-  3o = { (/) , { (/) } , { (/) , { (/) } } }