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Theorem elsucg 6376
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 6312 . . . 4 suc 𝐵 = (𝐵 ∪ {𝐵})
21eleq2i 2823 . . 3 (𝐴 ∈ suc 𝐵𝐴 ∈ (𝐵 ∪ {𝐵}))
3 elun 4100 . . 3 (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
42, 3bitri 275 . 2 (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
5 elsng 4587 . . 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 2111  cun 3895  {csn 4573  suc csuc 6308
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 2113  ax-9 2121  ax-ext 2703
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 2710  df-cleq 2723  df-clel 2806  df-v 3438  df-un 3902  df-sn 4574  df-suc 6312
This theorem is referenced by:  elsuc  6378  elelsuc  6381  sucidg  6389  ordsssuc  6397  ordsucelsuc  7752  suc11reg  9509  nlt1pi  10797
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