Metamath Proof Explorer


Theorem sspss

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

Ref Expression
Assertion sspss
|- ( A C_ B <-> ( A C. B \/ A = B ) )

Proof

Step Hyp Ref Expression
1 dfpss2
 |-  ( A C. B <-> ( A C_ B /\ -. A = B ) )
2 1 simplbi2
 |-  ( A C_ B -> ( -. A = B -> A C. B ) )
3 2 con1d
 |-  ( A C_ B -> ( -. A C. B -> A = B ) )
4 3 orrd
 |-  ( A C_ B -> ( A C. B \/ A = B ) )
5 pssss
 |-  ( A C. B -> A C_ B )
6 eqimss
 |-  ( A = B -> A C_ B )
7 5 6 jaoi
 |-  ( ( A C. B \/ A = B ) -> A C_ B )
8 4 7 impbii
 |-  ( A C_ B <-> ( A C. B \/ A = B ) )