Theorem indif1 3367

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 3366	. 2 |- ( B i^i ( A \ C ) ) = ( ( B i^i A ) \ C )
2		incom 3314	. 2 |- ( B i^i ( A \ C ) ) = ( ( A \ C ) i^i B )
3		incom 3314	. . 3 |- ( B i^i A ) = ( A i^i B )
4	3	difeq1i 3236	. 2 |- ( ( B i^i A ) \ C ) = ( ( A i^i B ) \ C )
5	1, 2, 4	3eqtr3i 2194	1 |- ( ( A \ C ) i^i B ) = ( ( A i^i B ) \ C )

Syntax hints:   = wceq 1343   \ cdif 3113   i^i cin 3115

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-rab 2453  df-v 2728  df-dif 3118  df-in 3122

This theorem is referenced by:  resdifcom  4902  resdmdfsn  4927