Theorem orddif 6460

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 6400	. 2 ⊢ (Ord 𝐴 → (𝐴 ∩ {𝐴}) = ∅)
2		disj3 4407	. . 3 ⊢ ((𝐴 ∩ {𝐴}) = ∅ ↔ 𝐴 = (𝐴 ∖ {𝐴}))
3		df-suc 6367	. . . . . 6 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
4	3	difeq1i 4070	. . . . 5 ⊢ (suc 𝐴 ∖ {𝐴}) = ((𝐴 ∪ {𝐴}) ∖ {𝐴})
5		difun2 4437	. . . . 5 ⊢ ((𝐴 ∪ {𝐴}) ∖ {𝐴}) = (𝐴 ∖ {𝐴})
6	4, 5	eqtri 2784	. . . 4 ⊢ (suc 𝐴 ∖ {𝐴}) = (𝐴 ∖ {𝐴})
7	6	eqeq2i 2774	. . 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 3896   ∪ cun 3897   ∩ cin 3898  ∅c0 4279  {csn 4584  Ord word 6360  suc csuc 6363

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 2733  ax-sep 5249  ax-pr 5391

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 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-eprel 5551  df-fr 5604  df-we 5606  df-ord 6364  df-suc 6367

This theorem is used by:  dif1enlem  9168  pssnn  9177  phplem2  9213  cantnfresb  44310