Metamath Proof Explorer

Theorem oveq2d

Description: Equality deduction for operation value. (Contributed by NM, 13-Mar-1995)

Ref Expression
Hypothesis oveq1d.1 
|- ( ph -> A = B )
Assertion oveq2d 
|- ( ph -> ( C F A ) = ( C F B ) )

Proof

Step Hyp Ref Expression
1 oveq1d.1 
 |-  ( ph -> A = B )
2 oveq2 
 |-  ( A = B -> ( C F A ) = ( C F B ) )
3 1 2 syl 
 |-  ( ph -> ( C F A ) = ( C F B ) )