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Theorem elsucg 6333
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 6272 . . . 4 suc 𝐵 = (𝐵 ∪ {𝐵})
21eleq2i 2830 . . 3 (𝐴 ∈ suc 𝐵𝐴 ∈ (𝐵 ∪ {𝐵}))
3 elun 4083 . . 3 (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
42, 3bitri 274 . 2 (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 ∈ {𝐵}))
5 elsng 4575 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
65orbi2d 913 . 2 (𝐴𝑉 → ((𝐴𝐵𝐴 ∈ {𝐵}) ↔ (𝐴𝐵𝐴 = 𝐵)))
74, 6bitrid 282 1 (𝐴𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wo 844   = wceq 1539  wcel 2106  cun 3885  {csn 4561  suc csuc 6268
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-un 3892  df-sn 4562  df-suc 6272
This theorem is referenced by:  elsuc  6335  elelsuc  6338  sucidg  6344  ordsssuc  6352  ordsucelsuc  7669  suc11reg  9377  nlt1pi  10662
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