Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  elsucg Structured version   Visualization version   GIF version

Theorem elsucg 6236
 Description: Membership in a successor. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 15-Sep-1995.)
Assertion
Ref Expression
elsucg (𝐴𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))

Proof of Theorem elsucg
StepHypRef Expression
1 df-suc 6175 . . . 4 suc 𝐵 = (𝐵 ∪ {𝐵})
21eleq2i 2843 . . 3 (𝐴 ∈ suc 𝐵𝐴 ∈ (𝐵 ∪ {𝐵}))
3 elun 4054 . . 3 (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
42, 3bitri 278 . 2 (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
5 elsng 4536 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
65orbi2d 913 . 2 (𝐴𝑉 → ((𝐴𝐵𝐴 ∈ {𝐵}) ↔ (𝐴𝐵𝐴 = 𝐵)))
74, 6syl5bb 286 1 (𝐴𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∨ wo 844   = wceq 1538   ∈ wcel 2111   ∪ cun 3856  {csn 4522  suc csuc 6171 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2729 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-v 3411  df-un 3863  df-sn 4523  df-suc 6175 This theorem is referenced by:  elsuc  6238  elelsuc  6241  sucidg  6247  ordsssuc  6255  ordsucelsuc  7536  suc11reg  9115  nlt1pi  10366
 Copyright terms: Public domain W3C validator