Metamath Proof Explorer


Theorem sscond

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

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

Proof

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