Metamath Proof Explorer


Theorem in13

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

Ref Expression
Assertion in13
|- ( A i^i ( B i^i C ) ) = ( C i^i ( B i^i A ) )

Proof

Step Hyp Ref Expression
1 in32
 |-  ( ( B i^i C ) i^i A ) = ( ( B i^i A ) i^i C )
2 incom
 |-  ( A i^i ( B i^i C ) ) = ( ( B i^i C ) i^i A )
3 incom
 |-  ( C i^i ( B i^i A ) ) = ( ( B i^i A ) i^i C )
4 1 2 3 3eqtr4i
 |-  ( A i^i ( B i^i C ) ) = ( C i^i ( B i^i A ) )