Theorem elsuci 4500

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 4468	. . . 4 ⊢ suc 𝐵 = (𝐵 ∪ {𝐵})
2	1	eleq2i 2298	. . 3 ⊢ (𝐴 ∈ suc 𝐵 ↔ 𝐴 ∈ (𝐵 ∪ {𝐵}))
3		elun 3348	. . 3 ⊢ (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
4	2, 3	bitri 184	. 2 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
5		elsni 3687	. . 3 ⊢ (𝐴 ∈ {𝐵} → 𝐴 = 𝐵)
6	5	orim2i 768	. 2 ⊢ ((𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}) → (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))
7	4, 6	sylbi 121	1 ⊢ (𝐴 ∈ suc 𝐵 → (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∨ wo 715   = wceq 1397   ∈ wcel 2202   ∪ cun 3198  {csn 3669  suc csuc 4462

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-un 3204  df-sn 3675  df-suc 4468

This theorem is referenced by:  trsucss  4520  onsucelsucexmid  4628  ordsoexmid  4660  ordsuc  4661  ordpwsucexmid  4668  nnsucelsuc  6658  nntri3or  6660  nnmordi  6683  nnaordex  6695  phplem3  7039  nninfninc  7321  nnnninf2  7325  3nelsucpw1  7451  3nsssucpw1  7453