Theorem df3o3 44456

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

Assertion

Ref	Expression
df3o3	⊢ 3o = {∅, {∅}, {∅, {∅}}}

Proof of Theorem df3o3

Step	Hyp	Ref	Expression
1		df-3o 8489	. 2 ⊢ 3o = suc 2o
2		df2o2 8496	. . . 4 ⊢ 2o = {∅, {∅}}
3	2	sneqi 4594	. . . 4 ⊢ {2o} = {{∅, {∅}}}
4	2, 3	uneq12i 4167	. . 3 ⊢ (2o ∪ {2o}) = ({∅, {∅}} ∪ {{∅, {∅}}})
5		df-suc 6374	. . 3 ⊢ suc 2o = (2o ∪ {2o})
6		df-tp 4592	. . 3 ⊢ {∅, {∅}, {∅, {∅}}} = ({∅, {∅}} ∪ {{∅, {∅}}})
7	4, 5, 6	3eqtr4i 2796	. 2 ⊢ suc 2o = {∅, {∅}, {∅, {∅}}}
8	1, 7	eqtri 2786	1 ⊢ 3o = {∅, {∅}, {∅, {∅}}}

This proof depends on syntax axioms:   = wceq 1569   ∪ cun 4151  ∅c0 4280  {csn 4587  {cpr 4589  {ctp 4591  suc csuc 6370  2_oc2o 8481  3_oc3o 8482

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3455  df-dif 4049  df-un 4153  df-nul 4281  df-sn 4588  df-pr 4590  df-tp 4592  df-suc 6374  df-1o 8487  df-2o 8488  df-3o 8489

This theorem is used by: (None)