Metamath Proof Explorer


Theorem ssdifd

Description: If A is contained in B , then ( A \ C ) is contained in ( B \ C ) . Deduction form of ssdif . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypothesis ssdifd.1
|- ( ph -> A C_ B )
Assertion ssdifd
|- ( ph -> ( A \ C ) C_ ( B \ C ) )

Proof

Step Hyp Ref Expression
1 ssdifd.1
 |-  ( ph -> A C_ B )
2 ssdif
 |-  ( A C_ B -> ( A \ C ) C_ ( B \ C ) )
3 1 2 syl
 |-  ( ph -> ( A \ C ) C_ ( B \ C ) )