Theorem orddif 6461

Description: Ordinal derived from its successor. (Contributed by NM, 20-May-1998.)

Assertion

Ref	Expression
orddif	⊢ (Ord 𝐴 → 𝐴 = (suc 𝐴 ∖ {𝐴}))

Proof of Theorem orddif

Step	Hyp	Ref	Expression
1		orddisj 6401	. 2 ⊢ (Ord 𝐴 → (𝐴 ∩ {𝐴}) = ∅)
2		disj3 4406	. . 3 ⊢ ((𝐴 ∩ {𝐴}) = ∅ ↔ 𝐴 = (𝐴 ∖ {𝐴}))
3		df-suc 6368	. . . . . 6 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
4	3	difeq1i 4059	. . . . 5 ⊢ (suc 𝐴 ∖ {𝐴}) = ((𝐴 ∪ {𝐴}) ∖ {𝐴})
5		difun2 4436	. . . . 5 ⊢ ((𝐴 ∪ {𝐴}) ∖ {𝐴}) = (𝐴 ∖ {𝐴})
6	4, 5	eqtri 2783	. . . 4 ⊢ (suc 𝐴 ∖ {𝐴}) = (𝐴 ∖ {𝐴})
7	6	eqeq2i 2773	. . 3 ⊢ (𝐴 = (suc 𝐴 ∖ {𝐴}) ↔ 𝐴 = (𝐴 ∖ {𝐴}))
8	2, 7	bitr4i 281	. 2 ⊢ ((𝐴 ∩ {𝐴}) = ∅ ↔ 𝐴 = (suc 𝐴 ∖ {𝐴}))
9	1, 8	sylib 221	1 ⊢ (Ord 𝐴 → 𝐴 = (suc 𝐴 ∖ {𝐴}))

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∖ cdif 4045   ∩ cin 4089   ∪ cun 4149  ∅c0 4278  {csn 4584  Ord word 6361  suc csuc 6364

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 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-pw 4558  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-eprel 5550  df-fr 5603  df-we 5605  df-ord 6365  df-suc 6368

This theorem is used by:  dif1enlem  9179  pssnn  9188  phplem2  9224  cantnfresb  44420