Metamath Proof Explorer


Theorem in4

Description: Rearrangement of intersection of 4 classes. (Contributed by NM, 21-Apr-2001)

Ref Expression
Assertion in4
|- ( ( A i^i B ) i^i ( C i^i D ) ) = ( ( A i^i C ) i^i ( B i^i D ) )

Proof

Step Hyp Ref Expression
1 in12
 |-  ( B i^i ( C i^i D ) ) = ( C i^i ( B i^i D ) )
2 1 ineq2i
 |-  ( A i^i ( B i^i ( C i^i D ) ) ) = ( A i^i ( C i^i ( B i^i D ) ) )
3 inass
 |-  ( ( A i^i B ) i^i ( C i^i D ) ) = ( A i^i ( B i^i ( C i^i D ) ) )
4 inass
 |-  ( ( A i^i C ) i^i ( B i^i D ) ) = ( A i^i ( C i^i ( B i^i D ) ) )
5 2 3 4 3eqtr4i
 |-  ( ( A i^i B ) i^i ( C i^i D ) ) = ( ( A i^i C ) i^i ( B i^i D ) )