Metamath Proof Explorer


Theorem in12

Description: A rearrangement of intersection. (Contributed by NM, 21-Apr-2001)

Ref Expression
Assertion in12
|- ( A i^i ( B i^i C ) ) = ( B i^i ( A i^i C ) )

Proof

Step Hyp Ref Expression
1 incom
 |-  ( A i^i B ) = ( B i^i A )
2 1 ineq1i
 |-  ( ( A i^i B ) i^i C ) = ( ( B i^i A ) i^i C )
3 inass
 |-  ( ( A i^i B ) i^i C ) = ( A i^i ( B i^i C ) )
4 inass
 |-  ( ( B i^i A ) i^i C ) = ( B i^i ( A i^i C ) )
5 2 3 4 3eqtr3i
 |-  ( A i^i ( B i^i C ) ) = ( B i^i ( A i^i C ) )