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Metamath Proof Explorer

Theorem ssdifsn

Description: Subset of a set with an element removed. (Contributed by Emmett Weisz, 7-Jul-2021) (Proof shortened by JJ, 31-May-2022)

Ref Expression
Assertion ssdifsn 
|- ( A C_ ( B \ { C } ) <-> ( A C_ B /\ -. C e. A ) )

Proof

Step Hyp Ref Expression
1 difss2 
 |-  ( A C_ ( B \ { C } ) -> A C_ B )
2 reldisj 
 |-  ( A C_ B -> ( ( A i^i { C } ) = (/) <-> A C_ ( B \ { C } ) ) )
3 2 bicomd 
 |-  ( A C_ B -> ( A C_ ( B \ { C } ) <-> ( A i^i { C } ) = (/) ) )
4 1 3 biadanii 
 |-  ( A C_ ( B \ { C } ) <-> ( A C_ B /\ ( A i^i { C } ) = (/) ) )
5 disjsn 
 |-  ( ( A i^i { C } ) = (/) <-> -. C e. A )
6 5 anbi2i 
 |-  ( ( A C_ B /\ ( A i^i { C } ) = (/) ) <-> ( A C_ B /\ -. C e. A ) )
7 4 6 bitri 
 |-  ( A C_ ( B \ { C } ) <-> ( A C_ B /\ -. C e. A ) )