Theorem indif2 3407

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 3373	. 2 |- ( ( A i^i B ) i^i ( _V \ C ) ) = ( A i^i ( B i^i ( _V \ C ) ) )
2		invdif 3405	. 2 |- ( ( A i^i B ) i^i ( _V \ C ) ) = ( ( A i^i B ) \ C )
3		invdif 3405	. . 3 |- ( B i^i ( _V \ C ) ) = ( B \ C )
4	3	ineq2i 3361	. 2 |- ( A i^i ( B i^i ( _V \ C ) ) ) = ( A i^i ( B \ C ) )
5	1, 2, 4	3eqtr3ri 2226	1 |- ( A i^i ( B \ C ) ) = ( ( A i^i B ) \ C )

Syntax hints:   = wceq 1364   _Vcvv 2763   \ cdif 3154   i^i cin 3156

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-dif 3159  df-in 3163

This theorem is referenced by:  indif1  3408  indifcom  3409  difopn  14344