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 ABBAB

Proof

Step Hyp Ref Expression
1 notnotb AB¬¬AB
2 difss BAB
3 2 biantrur BABBABBAB
4 difsnb ¬ABBA=B
5 4 necon3bbii ¬¬ABBAB
6 df-pss BABBABBAB
7 3 5 6 3bitr4i ¬¬ABBAB
8 1 7 bitri ABBAB