Theorem df3o3 44158

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 8460	. 2 ⊢ 3o = suc 2o
2		df2o2 8467	. . . 4 ⊢ 2o = {∅, {∅}}
3	2	sneqi 4595	. . . 4 ⊢ {2o} = {{∅, {∅}}}
4	2, 3	uneq12i 4113	. . 3 ⊢ (2o ∪ {2o}) = ({∅, {∅}} ∪ {{∅, {∅}}})
5		df-suc 6363	. . 3 ⊢ suc 2o = (2o ∪ {2o})
6		df-tp 4589	. . 3 ⊢ {∅, {∅}, {∅, {∅}}} = ({∅, {∅}} ∪ {{∅, {∅}}})
7	4, 5, 6	3eqtr4i 2793	. 2 ⊢ suc 2o = {∅, {∅}, {∅, {∅}}}
8	1, 7	eqtri 2783	1 ⊢ 3o = {∅, {∅}, {∅, {∅}}}

This proof depends on syntax axioms:   = wceq 1570   ∪ cun 3897  ∅c0 4279  {csn 4584  {cpr 4586  {ctp 4588  suc csuc 6359  2_oc2o 8452  3_oc3o 8453

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-dif 3902  df-un 3904  df-nul 4280  df-sn 4585  df-pr 4587  df-tp 4589  df-suc 6363  df-1o 8458  df-2o 8459  df-3o 8460

This theorem is used by: (None)