Theorem orddif 6487

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 6427	. 2 ⊢ (Ord 𝐴 → (𝐴 ∩ {𝐴}) = ∅)
2		disj3 4410	. . 3 ⊢ ((𝐴 ∩ {𝐴}) = ∅ ↔ 𝐴 = (𝐴 ∖ {𝐴}))
3		df-suc 6394	. . . . . 6 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
4	3	difeq1i 4062	. . . . 5 ⊢ (suc 𝐴 ∖ {𝐴}) = ((𝐴 ∪ {𝐴}) ∖ {𝐴})
5		difun2 4440	. . . . 5 ⊢ ((𝐴 ∪ {𝐴}) ∖ {𝐴}) = (𝐴 ∖ {𝐴})
6	4, 5	eqtri 2786	. . . 4 ⊢ (suc 𝐴 ∖ {𝐴}) = (𝐴 ∖ {𝐴})
7	6	eqeq2i 2776	. . 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 4048   ∩ cin 4092   ∪ cun 4152  ∅c0 4281  {csn 4588  Ord word 6387  suc csuc 6390

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5255  ax-pr 5398

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109  df-opab 5174  df-eprel 5558  df-fr 5611  df-we 5613  df-ord 6391  df-suc 6394

This theorem is used by:  dif1enlem  9213  pssnn  9222  phplem2  9258  cantnfresb  44587