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Theorem eldiftp 4059
Description: Membership in a set with three elements removed. Similar to eldifsn 4141 and eldifpr 4033. (Contributed by David A. Wheeler, 22-Jul-2017.)
Assertion
Ref Expression
eldiftp  |-  ( A  e.  ( B  \  { C ,  D ,  E } )  <->  ( A  e.  B  /\  ( A  =/=  C  /\  A  =/=  D  /\  A  =/= 
E ) ) )

Proof of Theorem eldiftp
StepHypRef Expression
1 eldif 3471 . 2  |-  ( A  e.  ( B  \  { C ,  D ,  E } )  <->  ( A  e.  B  /\  -.  A  e.  { C ,  D ,  E } ) )
2 eltpg 4058 . . . . 5  |-  ( A  e.  B  ->  ( A  e.  { C ,  D ,  E }  <->  ( A  =  C  \/  A  =  D  \/  A  =  E )
) )
32notbid 292 . . . 4  |-  ( A  e.  B  ->  ( -.  A  e.  { C ,  D ,  E }  <->  -.  ( A  =  C  \/  A  =  D  \/  A  =  E ) ) )
4 ne3anior 2780 . . . 4  |-  ( ( A  =/=  C  /\  A  =/=  D  /\  A  =/=  E )  <->  -.  ( A  =  C  \/  A  =  D  \/  A  =  E )
)
53, 4syl6bbr 263 . . 3  |-  ( A  e.  B  ->  ( -.  A  e.  { C ,  D ,  E }  <->  ( A  =/=  C  /\  A  =/=  D  /\  A  =/=  E ) ) )
65pm5.32i 635 . 2  |-  ( ( A  e.  B  /\  -.  A  e.  { C ,  D ,  E }
)  <->  ( A  e.  B  /\  ( A  =/=  C  /\  A  =/=  D  /\  A  =/= 
E ) ) )
71, 6bitri 249 1  |-  ( A  e.  ( B  \  { C ,  D ,  E } )  <->  ( A  e.  B  /\  ( A  =/=  C  /\  A  =/=  D  /\  A  =/= 
E ) ) )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    <-> wb 184    /\ wa 367    \/ w3o 970    /\ w3a 971    = wceq 1398    e. wcel 1823    =/= wne 2649    \ cdif 3458   {ctp 4020
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1623  ax-4 1636  ax-5 1709  ax-6 1752  ax-7 1795  ax-10 1842  ax-11 1847  ax-12 1859  ax-13 2004  ax-ext 2432
This theorem depends on definitions:  df-bi 185  df-or 368  df-an 369  df-3or 972  df-3an 973  df-tru 1401  df-ex 1618  df-nf 1622  df-sb 1745  df-clab 2440  df-cleq 2446  df-clel 2449  df-nfc 2604  df-ne 2651  df-v 3108  df-dif 3464  df-un 3466  df-sn 4017  df-pr 4019  df-tp 4021
This theorem is referenced by: (None)
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