Metamath Proof Explorer

Theorem in32

Description: A rearrangement of intersection. (Contributed by NM, 21-Apr-2001) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion in32 
|- ( ( A i^i B ) i^i C ) = ( ( A i^i C ) i^i B )

Proof

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