Metamath Proof Explorer

Theorem opeq2d

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

Ref Expression
Hypothesis opeq1d.1 
|- ( ph -> A = B )
Assertion opeq2d 
|- ( ph -> <. C , A >. = <. C , B >. )

Proof

Step Hyp Ref Expression
1 opeq1d.1 
 |-  ( ph -> A = B )
2 opeq2 
 |-  ( A = B -> <. C , A >. = <. C , B >. )
3 1 2 syl 
 |-  ( ph -> <. C , A >. = <. C , B >. )