Theorem neldif 2165

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 2057	. . . . 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 958   \ cdif 2044

This theorem is referenced by:  peano5 3153  clsval2 7685

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049

Copyright terms: Public domain