Theorem difcom 2349

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 2179	. . 3 |- (B u. C) = (C u. B)
2	1	sseq2i 2089	. 2 |- (A (_ (B u. C) <-> A (_ (C u. B))
3		ssundif 2348	. 2 |- (A (_ (B u. C) <-> (A \ B) (_ C)
4		ssundif 2348	. 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 2047   u. cun 2048   (_ wss 2050

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-v 1815  df-dif 2052  df-un 2053  df-in 2054  df-ss 2056

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