Metamath Proof Explorer


Theorem difsnb

Description: ( B \ { A } ) equals B if and only if A is not a member of B . Generalization of difsn . (Contributed by David Moews, 1-May-2017)

Ref Expression
Assertion difsnb
|- ( -. A e. B <-> ( B \ { A } ) = B )

Proof

Step Hyp Ref Expression
1 difsn
 |-  ( -. A e. B -> ( B \ { A } ) = B )
2 neldifsnd
 |-  ( A e. B -> -. A e. ( B \ { A } ) )
3 nelne1
 |-  ( ( A e. B /\ -. A e. ( B \ { A } ) ) -> B =/= ( B \ { A } ) )
4 2 3 mpdan
 |-  ( A e. B -> B =/= ( B \ { A } ) )
5 4 necomd
 |-  ( A e. B -> ( B \ { A } ) =/= B )
6 5 necon2bi
 |-  ( ( B \ { A } ) = B -> -. A e. B )
7 1 6 impbii
 |-  ( -. A e. B <-> ( B \ { A } ) = B )