Theorem difcom 2345

Description: Swap the arguments of a class difference.

Assertion

Ref	Expression
difcom	|- ((A \ B) (_ C <-> (A \ C) (_ B)

Proof of Theorem difcom

Step	Hyp	Ref	Expression
1		uncom 2176	. . 3 |- (B u. C) = (C u. B)
2	1	sseq2i 2086	. 2 |- (A (_ (B u. C) <-> A (_ (C u. B))
3		ssundif 2344	. 2 |- (A (_ (B u. C) <-> (A \ B) (_ C)
4		ssundif 2344	. 2 |- (A (_ (C u. B) <-> (A \ C) (_ B)
5	2, 3, 4	3bitr3 181	1 |- ((A \ B) (_ C <-> (A \ C) (_ B)

Syntax hints:   <-> wb 146   \ cdif 2044   u. cun 2045   (_ wss 2047

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-or 224  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049  df-un 2050  df-in 2051  df-ss 2053

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