Metamath Proof Explorer

Theorem oveqd

Description: Equality deduction for operation value. (Contributed by NM, 9-Sep-2006)

Ref Expression
Hypothesis oveq1d.1 
|- ( ph -> A = B )
Assertion oveqd 
|- ( ph -> ( C A D ) = ( C B D ) )

Proof

Step Hyp Ref Expression
1 oveq1d.1 
 |-  ( ph -> A = B )
2 oveq 
 |-  ( A = B -> ( C A D ) = ( C B D ) )
3 1 2 syl 
 |-  ( ph -> ( C A D ) = ( C B D ) )