Theorem neldif 2162

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

Step	Hyp	Ref	Expression
1		eldif 2054	. . . . 5 |- (A e. (B \ C) <-> (A e. B /\ -. A e. C))
2	1	biimpr 152	. . . 4 |- ((A e. B /\ -. A e. C) -> A e. (B \ C))
3	2	ex 373	. . 3 |- (A e. B -> (-. A e. C -> A e. (B \ C)))
4	3	con1d 93	. 2 |- (A e. B -> (-. A e. (B \ C) -> A e. C))
5	4	imp 350	1 |- ((A e. B /\ -. A e. (B \ C)) -> A e. C)

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   e. wcel 957   \ cdif 2041

This theorem is referenced by:  peano5 3149  clsval2 7645

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217  ax-ext 1458

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-sb 1171  df-clab 1463  df-cleq 1468  df-clel 1471  df-v 1809  df-dif 2046

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