Metamath Proof Explorer


Theorem difeq12d

Description: Equality deduction for class difference. (Contributed by FL, 29-May-2014)

Ref Expression
Hypotheses difeq12d.1
|- ( ph -> A = B )
difeq12d.2
|- ( ph -> C = D )
Assertion difeq12d
|- ( ph -> ( A \ C ) = ( B \ D ) )

Proof

Step Hyp Ref Expression
1 difeq12d.1
 |-  ( ph -> A = B )
2 difeq12d.2
 |-  ( ph -> C = D )
3 1 difeq1d
 |-  ( ph -> ( A \ C ) = ( B \ C ) )
4 2 difeq2d
 |-  ( ph -> ( B \ C ) = ( B \ D ) )
5 3 4 eqtrd
 |-  ( ph -> ( A \ C ) = ( B \ D ) )