Metamath Proof Explorer

Theorem xpeq2d

Description: Equality deduction for Cartesian product. (Contributed by Jeff Madsen, 17-Jun-2010)

Ref Expression
Hypothesis xpeq1d.1 
|- ( ph -> A = B )
Assertion xpeq2d 
|- ( ph -> ( C X. A ) = ( C X. B ) )

Proof

Step Hyp Ref Expression
1 xpeq1d.1 
 |-  ( ph -> A = B )
2 xpeq2 
 |-  ( A = B -> ( C X. A ) = ( C X. B ) )
3 1 2 syl 
 |-  ( ph -> ( C X. A ) = ( C X. B ) )