Theorem indif 3498

Description: Intersection with class difference. Theorem 34 of [Suppes] p. 29. (Contributed by NM, 17-Aug-2004.)

Assertion

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

Proof of Theorem indif

Step	Hyp	Ref	Expression
1		dfin4 3496	. 2 |- (A i^i (A \ B)) = (A \ (A \ (A \ B)))
2		dfin4 3496	. . 3 |- (A i^i B) = (A \ (A \ B))
3	2	difeq2i 3383	. 2 |- (A \ (A i^i B)) = (A \ (A \ (A \ B)))
4		difin 3493	. 2 |- (A \ (A i^i B)) = (A \ B)
5	1, 3, 4	3eqtr2i 2379	1 |- (A i^i (A \ B)) = (A \ B)

Syntax hints:   = wceq 1642   \ cdif 3207   i^i cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260

This theorem is referenced by: (None)