Metamath Proof Explorer


Theorem indifcom

Description: Commutation law for intersection and difference. (Contributed by Scott Fenton, 18-Feb-2013)

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

Proof

Step Hyp Ref Expression
1 incom
 |-  ( A i^i B ) = ( B i^i A )
2 1 difeq1i
 |-  ( ( A i^i B ) \ C ) = ( ( B i^i A ) \ C )
3 indif2
 |-  ( A i^i ( B \ C ) ) = ( ( A i^i B ) \ C )
4 indif2
 |-  ( B i^i ( A \ C ) ) = ( ( B i^i A ) \ C )
5 2 3 4 3eqtr4i
 |-  ( A i^i ( B \ C ) ) = ( B i^i ( A \ C ) )