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 B B A B

Proof

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