Metamath Proof Explorer

Theorem eldifsni

Description: Membership in a set with an element removed. (Contributed by NM, 10-Mar-2015)

Ref Expression
Assertion eldifsni 
|- ( A e. ( B \ { C } ) -> A =/= C )

Proof

Step Hyp Ref Expression
1 eldifsn 
 |-  ( A e. ( B \ { C } ) <-> ( A e. B /\ A =/= C ) )
2 1 simprbi 
 |-  ( A e. ( B \ { C } ) -> A =/= C )