Theorem difun1 1687

Description: A relationship involving double difference and union.

Assertion

Ref	Expression
difun1	|- (A \ (B u. C)) = ((A \ B) \ C)

Proof of Theorem difun1

Step	Hyp	Ref	Expression
1		inass 1650	. . 3 |- ((A i^i (V \ B)) i^i (V \ C)) = (A i^i ((V \ B) i^i (V \ C)))
2		invdif 1674	. . 3 |- ((A i^i (V \ B)) i^i (V \ C)) = ((A i^i (V \ B)) \ C)
3		undm 1685	. . . . 5 |- (V \ (B u. C)) = ((V \ B) i^i (V \ C))
4	3	ineq2i 1642	. . . 4 |- (A i^i (V \ (B u. C))) = (A i^i ((V \ B) i^i (V \ C)))
5		invdif 1674	. . . 4 |- (A i^i (V \ (B u. C))) = (A \ (B u. C))
6	4, 5	eqtr3 1121	. . 3 |- (A i^i ((V \ B) i^i (V \ C))) = (A \ (B u. C))
7	1, 2, 6	3eqtr3 1124	. 2 |- ((A i^i (V \ B)) \ C) = (A \ (B u. C))
8		invdif 1674	. . 3 |- (A i^i (V \ B)) = (A \ B)
9	8	difeq1i 1584	. 2 |- ((A i^i (V \ B)) \ C) = ((A \ B) \ C)
10	7, 9	eqtr3 1121	1 |- (A \ (B u. C)) = ((A \ B) \ C)

Syntax hints:   = wceq 1091  Vcvv 1348   \ cdif 1484   u. cun 1485   i^i cin 1486

This theorem is referenced by:  dif23 1688

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-un 1490  df-in 1491

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