Theorem indif 1675

Description: Intersection with class difference. Theorem 34 of [Suppes] p. 29.

Assertion

Ref	Expression
indif	|- (A i^i (A \ B)) = (A \ B)

Proof of Theorem indif

Step	Hyp	Ref	Expression
1		dfin4 1673	. 2 |- (A i^i (A \ B)) = (A \ (A \ (A \ B)))
2		dfin4 1673	. . . 4 |- (A i^i B) = (A \ (A \ B))
3	2	difeq2i 1585	. . 3 |- (A \ (A i^i B)) = (A \ (A \ (A \ B)))
4		difin 1670	. . 3 |- (A \ (A i^i B)) = (A \ B)
5	3, 4	eqtr3 1121	. 2 |- (A \ (A \ (A \ B))) = (A \ B)
6	1, 5	eqtr 1119	1 |- (A i^i (A \ B)) = (A \ B)

Syntax hints:   = wceq 1091   \ cdif 1484   i^i cin 1486

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-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-in 1491  df-ss 1492

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