Theorem orddif 6463

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 6403	. 2 ⊢ (Ord 𝐴 → (𝐴 ∩ {𝐴}) = ∅)
2		disj3 4414	. . 3 ⊢ ((𝐴 ∩ {𝐴}) = ∅ ↔ 𝐴 = (𝐴 ∖ {𝐴}))
3		df-suc 6370	. . . . . 6 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
4	3	difeq1i 4077	. . . . 5 ⊢ (suc 𝐴 ∖ {𝐴}) = ((𝐴 ∪ {𝐴}) ∖ {𝐴})
5		difun2 4444	. . . . 5 ⊢ ((𝐴 ∪ {𝐴}) ∖ {𝐴}) = (𝐴 ∖ {𝐴})
6	4, 5	eqtri 2788	. . . 4 ⊢ (suc 𝐴 ∖ {𝐴}) = (𝐴 ∖ {𝐴})
7	6	eqeq2i 2778	. . 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 1570   ∖ cdif 3903   ∪ cun 3904   ∩ cin 3905  ∅c0 4286  {csn 4591  Ord word 6363  suc csuc 6366

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 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-pw 4566  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-eprel 5563  df-fr 5616  df-we 5618  df-ord 6367  df-suc 6370

This theorem is used by:  dif1enlem  9147  pssnn  9156  phplem2  9192  cantnfresb  44084