Metamath Proof Explorer

Theorem opeq1d

Description: Equality deduction for ordered pairs. (Contributed by NM, 16-Dec-2006)

Ref Expression
Hypothesis opeq1d.1 
|- ( ph -> A = B )
Assertion opeq1d 
|- ( ph -> <. A , C >. = <. B , C >. )

Proof

Step Hyp Ref Expression
1 opeq1d.1 
 |-  ( ph -> A = B )
2 opeq1 
 |-  ( A = B -> <. A , C >. = <. B , C >. )
3 1 2 syl 
 |-  ( ph -> <. A , C >. = <. B , C >. )