Metamath Proof Explorer


Theorem npss0

Description: No set is a proper subset of the empty set. (Contributed by NM, 17-Jun-1998) (Proof shortened by Andrew Salmon, 26-Jun-2011) (Proof shortened by JJ, 14-Jul-2021)

Ref Expression
Assertion npss0
|- -. A C. (/)

Proof

Step Hyp Ref Expression
1 0ss
 |-  (/) C_ A
2 ssnpss
 |-  ( (/) C_ A -> -. A C. (/) )
3 1 2 ax-mp
 |-  -. A C. (/)