Metamath Proof Explorer

Theorem inxp

Description: Intersection of two Cartesian products. Exercise 9 of TakeutiZaring p. 25. (Contributed by NM, 3-Aug-1994) (Proof shortened by Andrew Salmon, 27-Aug-2011)

Ref Expression
Assertion inxp 
|- ( ( A X. B ) i^i ( C X. D ) ) = ( ( A i^i C ) X. ( B i^i D ) )

Proof

Step Hyp Ref Expression
1 inopab 
 |-  ( { <. x , y >. | ( x e. A /\ y e. B ) } i^i { <. x , y >. | ( x e. C /\ y e. D ) } ) = { <. x , y >. | ( ( x e. A /\ y e. B ) /\ ( x e. C /\ y e. D ) ) }
2 an4 
 |-  ( ( ( x e. A /\ y e. B ) /\ ( x e. C /\ y e. D ) ) <-> ( ( x e. A /\ x e. C ) /\ ( y e. B /\ y e. D ) ) )
3 elin 
 |-  ( x e. ( A i^i C ) <-> ( x e. A /\ x e. C ) )
4 elin 
 |-  ( y e. ( B i^i D ) <-> ( y e. B /\ y e. D ) )
5 3 4 anbi12i 
 |-  ( ( x e. ( A i^i C ) /\ y e. ( B i^i D ) ) <-> ( ( x e. A /\ x e. C ) /\ ( y e. B /\ y e. D ) ) )
6 2 5 bitr4i 
 |-  ( ( ( x e. A /\ y e. B ) /\ ( x e. C /\ y e. D ) ) <-> ( x e. ( A i^i C ) /\ y e. ( B i^i D ) ) )
7 6 opabbii 
 |-  { <. x , y >. | ( ( x e. A /\ y e. B ) /\ ( x e. C /\ y e. D ) ) } = { <. x , y >. | ( x e. ( A i^i C ) /\ y e. ( B i^i D ) ) }
8 1 7 eqtri 
 |-  ( { <. x , y >. | ( x e. A /\ y e. B ) } i^i { <. x , y >. | ( x e. C /\ y e. D ) } ) = { <. x , y >. | ( x e. ( A i^i C ) /\ y e. ( B i^i D ) ) }
9 df-xp 
 |-  ( A X. B ) = { <. x , y >. | ( x e. A /\ y e. B ) }
10 df-xp 
 |-  ( C X. D ) = { <. x , y >. | ( x e. C /\ y e. D ) }
11 9 10 ineq12i 
 |-  ( ( A X. B ) i^i ( C X. D ) ) = ( { <. x , y >. | ( x e. A /\ y e. B ) } i^i { <. x , y >. | ( x e. C /\ y e. D ) } )
12 df-xp 
 |-  ( ( A i^i C ) X. ( B i^i D ) ) = { <. x , y >. | ( x e. ( A i^i C ) /\ y e. ( B i^i D ) ) }
13 8 11 12 3eqtr4i 
 |-  ( ( A X. B ) i^i ( C X. D ) ) = ( ( A i^i C ) X. ( B i^i D ) )