Metamath Proof Explorer


Theorem xpeq12d

Description: Equality deduction for Cartesian product. (Contributed by NM, 8-Dec-2013)

Ref Expression
Hypotheses xpeq1d.1
|- ( ph -> A = B )
xpeq12d.2
|- ( ph -> C = D )
Assertion xpeq12d
|- ( ph -> ( A X. C ) = ( B X. D ) )

Proof

Step Hyp Ref Expression
1 xpeq1d.1
 |-  ( ph -> A = B )
2 xpeq12d.2
 |-  ( ph -> C = D )
3 xpeq12
 |-  ( ( A = B /\ C = D ) -> ( A X. C ) = ( B X. D ) )
4 1 2 3 syl2anc
 |-  ( ph -> ( A X. C ) = ( B X. D ) )