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Theorem elsuci 4405
Description: Membership in a successor. This one-way implication does not require that either  A or  B be sets. (Contributed by NM, 6-Jun-1994.)
Assertion
Ref Expression
elsuci  |-  ( A  e.  suc  B  -> 
( A  e.  B  \/  A  =  B
) )

Proof of Theorem elsuci
StepHypRef Expression
1 df-suc 4373 . . . 4  |-  suc  B  =  ( B  u.  { B } )
21eleq2i 2244 . . 3  |-  ( A  e.  suc  B  <->  A  e.  ( B  u.  { B } ) )
3 elun 3278 . . 3  |-  ( A  e.  ( B  u.  { B } )  <->  ( A  e.  B  \/  A  e.  { B } ) )
42, 3bitri 184 . 2  |-  ( A  e.  suc  B  <->  ( A  e.  B  \/  A  e.  { B } ) )
5 elsni 3612 . . 3  |-  ( A  e.  { B }  ->  A  =  B )
65orim2i 761 . 2  |-  ( ( A  e.  B  \/  A  e.  { B } )  ->  ( A  e.  B  \/  A  =  B )
)
74, 6sylbi 121 1  |-  ( A  e.  suc  B  -> 
( A  e.  B  \/  A  =  B
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    \/ wo 708    = wceq 1353    e. wcel 2148    u. cun 3129   {csn 3594   suc csuc 4367
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-un 3135  df-sn 3600  df-suc 4373
This theorem is referenced by:  trsucss  4425  onsucelsucexmid  4531  ordsoexmid  4563  ordsuc  4564  ordpwsucexmid  4571  nnsucelsuc  6494  nntri3or  6496  nnmordi  6519  nnaordex  6531  phplem3  6856  nnnninf2  7127  3nelsucpw1  7235  3nsssucpw1  7237
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