Metamath Proof Explorer


Theorem altxpeq1

Description: Equality for alternate Cartesian products. (Contributed by Scott Fenton, 24-Mar-2012)

Ref Expression
Assertion altxpeq1
|- ( A = B -> ( A XX. C ) = ( B XX. C ) )

Proof

Step Hyp Ref Expression
1 rexeq
 |-  ( A = B -> ( E. x e. A E. y e. C z = << x , y >> <-> E. x e. B E. y e. C z = << x , y >> ) )
2 1 abbidv
 |-  ( A = B -> { z | E. x e. A E. y e. C z = << x , y >> } = { z | E. x e. B E. y e. C z = << x , y >> } )
3 df-altxp
 |-  ( A XX. C ) = { z | E. x e. A E. y e. C z = << x , y >> }
4 df-altxp
 |-  ( B XX. C ) = { z | E. x e. B E. y e. C z = << x , y >> }
5 2 3 4 3eqtr4g
 |-  ( A = B -> ( A XX. C ) = ( B XX. C ) )