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Theorem elsucg 6395
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 6331 . . . 4 suc 𝐵 = (𝐵 ∪ {𝐵})
21eleq2i 2829 . . 3 (𝐴 ∈ suc 𝐵𝐴 ∈ (𝐵 ∪ {𝐵}))
3 elun 4107 . . 3 (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
42, 3bitri 275 . 2 (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
5 elsng 4596 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
65orbi2d 916 . 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 848   = wceq 1542  wcel 2114  cun 3901  {csn 4582  suc csuc 6327
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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3444  df-un 3908  df-sn 4583  df-suc 6331
This theorem is referenced by:  elsuc  6397  elelsuc  6400  sucidg  6408  ordsssuc  6416  ordsucelsuc  7774  suc11reg  9540  nlt1pi  10829
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