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Theorem disjel 3501
Description: A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)
Assertion
Ref Expression
disjel  |-  ( ( ( A  i^i  B
)  =  (/)  /\  C  e.  A )  ->  -.  C  e.  B )

Proof of Theorem disjel
StepHypRef Expression
1 disj3 3499 . . 3  |-  ( ( A  i^i  B )  =  (/)  <->  A  =  ( A  \  B ) )
2 eleq2 2344 . . . 4  |-  ( A  =  ( A  \  B )  ->  ( C  e.  A  <->  C  e.  ( A  \  B ) ) )
3 eldifn 3299 . . . 4  |-  ( C  e.  ( A  \  B )  ->  -.  C  e.  B )
42, 3syl6bi 219 . . 3  |-  ( A  =  ( A  \  B )  ->  ( C  e.  A  ->  -.  C  e.  B ) )
51, 4sylbi 187 . 2  |-  ( ( A  i^i  B )  =  (/)  ->  ( C  e.  A  ->  -.  C  e.  B )
)
65imp 418 1  |-  ( ( ( A  i^i  B
)  =  (/)  /\  C  e.  A )  ->  -.  C  e.  B )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    /\ wa 358    = wceq 1623    e. wcel 1684    \ cdif 3149    i^i cin 3151   (/)c0 3455
This theorem is referenced by:  disjxun  4021  fvun1  5590  dedekindle  24083
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866  ax-ext 2264
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630  df-clab 2270  df-cleq 2276  df-clel 2279  df-nfc 2408  df-ral 2548  df-v 2790  df-dif 3155  df-in 3159  df-nul 3456
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