Theorem indif2 3498

Description: Bring an intersection in and out of a class difference. (Contributed by Jeff Hankins, 15-Jul-2009.)

Assertion

Ref	Expression
indif2	|- (A i^i (B \ C)) = ((A i^i B) \ C)

Proof of Theorem indif2

Step	Hyp	Ref	Expression
1		inass 3465	. 2 |- ((A i^i B) i^i (_V \ C)) = (A i^i (B i^i (_V \ C)))
2		invdif 3496	. 2 |- ((A i^i B) i^i (_V \ C)) = ((A i^i B) \ C)
3		invdif 3496	. . 3 |- (B i^i (_V \ C)) = (B \ C)
4	3	ineq2i 3454	. 2 |- (A i^i (B i^i (_V \ C))) = (A i^i (B \ C))
5	1, 2, 4	3eqtr3ri 2382	1 |- (A i^i (B \ C)) = ((A i^i B) \ C)

Syntax hints:   = wceq 1642  _Vcvv 2859   \ cdif 3206   i^i cin 3208

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 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215

This theorem is referenced by:  indif1  3499  indifcom  3500