Metamath Proof Explorer

Theorem difeq2d

Description: Deduction adding difference to the left in a class equality. (Contributed by NM, 15-Nov-2002)

Ref Expression
Hypothesis difeq1d.1 
|- ( ph -> A = B )
Assertion difeq2d 
|- ( ph -> ( C \ A ) = ( C \ B ) )

Proof

Step Hyp Ref Expression
1 difeq1d.1 
 |-  ( ph -> A = B )
2 difeq2 
 |-  ( A = B -> ( C \ A ) = ( C \ B ) )
3 1 2 syl 
 |-  ( ph -> ( C \ A ) = ( C \ B ) )