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Theorem neldif 2155
Description: Implication of membership in a class difference.
Assertion
Ref Expression
neldif |- ((A e. B /\ -. A e. (B \ C)) -> A e. C)
Proof of Theorem neldif
StepHypRef Expression
1 eldif 2047 . . . . 5 |- (A e. (B \ C) <-> (A e. B /\ -. A e. C))
21biimpr 152 . . . 4 |- ((A e. B /\ -. A e. C) -> A e. (B \ C))
32ex 373 . . 3 |- (A e. B -> (-. A e. C -> A e. (B \ C)))
43con1d 93 . 2 |- (A e. B -> (-. A e. (B \ C) -> A e. C))
54imp 350 1 |- ((A e. B /\ -. A e. (B \ C)) -> A e. C)
Colors of variables: wff set class
Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   e. wcel 955   \ cdif 2034
This theorem is referenced by:  peano5 3143  clsval2 7627
This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452
This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-v 1803  df-dif 2039
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