Metamath Proof Explorer

Theorem xpeq1d

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

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

Proof

Step Hyp Ref Expression
1 xpeq1d.1 
 |-  ( ph -> A = B )
2 xpeq1 
 |-  ( A = B -> ( A X. C ) = ( B X. C ) )
3 1 2 syl 
 |-  ( ph -> ( A X. C ) = ( B X. C ) )