Metamath Proof Explorer


Theorem indif2

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

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

Proof

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