Metamath Proof Explorer


Theorem difsnpss

Description: ( B \ { A } ) is a proper subclass of B if and only if A is a member of B . (Contributed by David Moews, 1-May-2017)

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

Proof

Step Hyp Ref Expression
1 notnotb
 |-  ( A e. B <-> -. -. A e. B )
2 difss
 |-  ( B \ { A } ) C_ B
3 2 biantrur
 |-  ( ( B \ { A } ) =/= B <-> ( ( B \ { A } ) C_ B /\ ( B \ { A } ) =/= B ) )
4 difsnb
 |-  ( -. A e. B <-> ( B \ { A } ) = B )
5 4 necon3bbii
 |-  ( -. -. A e. B <-> ( B \ { A } ) =/= B )
6 df-pss
 |-  ( ( B \ { A } ) C. B <-> ( ( B \ { A } ) C_ B /\ ( B \ { A } ) =/= B ) )
7 3 5 6 3bitr4i
 |-  ( -. -. A e. B <-> ( B \ { A } ) C. B )
8 1 7 bitri
 |-  ( A e. B <-> ( B \ { A } ) C. B )