Theorem indif1 3500

Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)

Assertion

Ref	Expression
indif1	|- ((A \ C) i^i B) = ((A i^i B) \ C)

Proof of Theorem indif1

Step	Hyp	Ref	Expression
1		indif2 3499	. 2 |- (B i^i (A \ C)) = ((B i^i A) \ C)
2		incom 3449	. 2 |- (B i^i (A \ C)) = ((A \ C) i^i B)
3		incom 3449	. . 3 |- (B i^i A) = (A i^i B)
4	3	difeq1i 3382	. 2 |- ((B i^i A) \ C) = ((A i^i B) \ C)
5	1, 2, 4	3eqtr3i 2381	1 |- ((A \ C) i^i B) = ((A i^i B) \ C)

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

This theorem is referenced by: (None)