Theorem elsuci 4089

Description: Membership in a successor. This one-way implication does not require that either A or B be sets. (Contributed by NM, 6-Jun-1994.)

Assertion

Ref	Expression
elsuci	⊢ (A ∈ suc B → (A ∈ B ∨ A = B))

Proof of Theorem elsuci

Step	Hyp	Ref	Expression
1		df-suc 4057	. . . 4 ⊢ suc B = (B ∪ {B})
2	1	eleq2i 2086	. . 3 ⊢ (A ∈ suc B ↔ A ∈ (B ∪ {B}))
3		elun 3061	. . 3 ⊢ (A ∈ (B ∪ {B}) ↔ (A ∈ B ∨ A ∈ {B}))
4	2, 3	bitri 173	. 2 ⊢ (A ∈ suc B ↔ (A ∈ B ∨ A ∈ {B}))
5		elsni 3374	. . 3 ⊢ (A ∈ {B} → A = B)
6	5	orim2i 665	. 2 ⊢ ((A ∈ B ∨ A ∈ {B}) → (A ∈ B ∨ A = B))
7	4, 6	sylbi 114	1 ⊢ (A ∈ suc B → (A ∈ B ∨ A = B))

Syntax hints:   → wi 4   ∨ wo 616   = wceq 1228   ∈ wcel 1374   ∪ cun 2892  {csn 3350  suc csuc 4051

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537  df-un 2899  df-sn 3356  df-suc 4057

This theorem is referenced by:  suctr  4108  trsucss  4110  onsucelsucexmid  4199  ordsoexmid  4224  ordsuc  4225  ordpwsucexmid  4230  nnsucelsuc  5985  nntri3or  5987  nnmordi  6000  nnaordex  6011