Metamath Proof Explorer


Theorem sspss

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

Ref Expression
Assertion sspss ( 𝐴𝐵 ↔ ( 𝐴𝐵𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 dfpss2 ( 𝐴𝐵 ↔ ( 𝐴𝐵 ∧ ¬ 𝐴 = 𝐵 ) )
2 1 simplbi2 ( 𝐴𝐵 → ( ¬ 𝐴 = 𝐵𝐴𝐵 ) )
3 2 con1d ( 𝐴𝐵 → ( ¬ 𝐴𝐵𝐴 = 𝐵 ) )
4 3 orrd ( 𝐴𝐵 → ( 𝐴𝐵𝐴 = 𝐵 ) )
5 pssss ( 𝐴𝐵𝐴𝐵 )
6 eqimss ( 𝐴 = 𝐵𝐴𝐵 )
7 5 6 jaoi ( ( 𝐴𝐵𝐴 = 𝐵 ) → 𝐴𝐵 )
8 4 7 impbii ( 𝐴𝐵 ↔ ( 𝐴𝐵𝐴 = 𝐵 ) )