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Theorem eldiftp 4006
Description: Membership in a set with three elements removed. Similar to eldifsn 4088 and eldifpr 3982. (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 3400 . 2  |-  ( A  e.  ( B  \  { C ,  D ,  E } )  <->  ( A  e.  B  /\  -.  A  e.  { C ,  D ,  E } ) )
2 eltpg 4005 . . . . 5  |-  ( A  e.  B  ->  ( A  e.  { C ,  D ,  E }  <->  ( A  =  C  \/  A  =  D  \/  A  =  E )
) )
32notbid 301 . . . 4  |-  ( A  e.  B  ->  ( -.  A  e.  { C ,  D ,  E }  <->  -.  ( A  =  C  \/  A  =  D  \/  A  =  E ) ) )
4 ne3anior 2736 . . . 4  |-  ( ( A  =/=  C  /\  A  =/=  D  /\  A  =/=  E )  <->  -.  ( A  =  C  \/  A  =  D  \/  A  =  E )
)
53, 4syl6bbr 271 . . 3  |-  ( A  e.  B  ->  ( -.  A  e.  { C ,  D ,  E }  <->  ( A  =/=  C  /\  A  =/=  D  /\  A  =/=  E ) ) )
65pm5.32i 649 . 2  |-  ( ( A  e.  B  /\  -.  A  e.  { C ,  D ,  E }
)  <->  ( A  e.  B  /\  ( A  =/=  C  /\  A  =/=  D  /\  A  =/= 
E ) ) )
71, 6bitri 257 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 189    /\ wa 376    \/ w3o 1006    /\ w3a 1007    = wceq 1452    e. wcel 1904    =/= wne 2641    \ cdif 3387   {ctp 3963
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-10 1932  ax-11 1937  ax-12 1950  ax-13 2104  ax-ext 2451
This theorem depends on definitions:  df-bi 190  df-or 377  df-an 378  df-3or 1008  df-3an 1009  df-tru 1455  df-ex 1672  df-nf 1676  df-sb 1806  df-clab 2458  df-cleq 2464  df-clel 2467  df-nfc 2601  df-ne 2643  df-v 3033  df-dif 3393  df-un 3395  df-sn 3960  df-pr 3962  df-tp 3964
This theorem is referenced by: (None)
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