Metamath Proof Explorer


Theorem sspss

Description: Subclass in terms of proper subclass. (Contributed by NM, 25-Feb-1996)

Ref Expression
Assertion sspss A B A B A = B

Proof

Step Hyp Ref Expression
1 dfpss2 A B A B ¬ A = B
2 1 simplbi2 A B ¬ A = B A B
3 2 con1d A B ¬ A B A = B
4 3 orrd A B A B A = B
5 pssss A B A B
6 eqimss A = B A B
7 5 6 jaoi A B A = B A B
8 4 7 impbii A B A B A = B