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

Theorem elsucg 6385
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 6321 . . . 4 suc 𝐵 = (𝐵 ∪ {𝐵})
21eleq2i 2826 . . 3 (𝐴 ∈ suc 𝐵𝐴 ∈ (𝐵 ∪ {𝐵}))
3 elun 4103 . . 3 (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
42, 3bitri 275 . 2 (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
5 elsng 4592 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
65orbi2d 915 . 2 (𝐴𝑉 → ((𝐴𝐵𝐴 ∈ {𝐵}) ↔ (𝐴𝐵𝐴 = 𝐵)))
74, 6bitrid 283 1 (𝐴𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wo 847   = wceq 1541  wcel 2113  cun 3897  {csn 4578  suc csuc 6317
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2713  df-cleq 2726  df-clel 2809  df-v 3440  df-un 3904  df-sn 4579  df-suc 6321
This theorem is referenced by:  elsuc  6387  elelsuc  6390  sucidg  6398  ordsssuc  6406  ordsucelsuc  7762  suc11reg  9526  nlt1pi  10815
  Copyright terms: Public domain W3C validator