Metamath Proof Explorer


Theorem in31

Description: A rearrangement of intersection. (Contributed by NM, 27-Aug-2012)

Ref Expression
Assertion in31
|- ( ( A i^i B ) i^i C ) = ( ( C i^i B ) i^i A )

Proof

Step Hyp Ref Expression
1 in12
 |-  ( C i^i ( A i^i B ) ) = ( A i^i ( C i^i B ) )
2 incom
 |-  ( ( A i^i B ) i^i C ) = ( C i^i ( A i^i B ) )
3 incom
 |-  ( ( C i^i B ) i^i A ) = ( A i^i ( C i^i B ) )
4 1 2 3 3eqtr4i
 |-  ( ( A i^i B ) i^i C ) = ( ( C i^i B ) i^i A )