Metamath Proof Explorer


Theorem ab0

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 (from which it could be proved using as many essential proof steps but one fewer syntactic step, at the cost of depending on df-ne ). (Contributed by BJ, 19-Mar-2021)

Ref Expression
Assertion ab0
|- ( { x | ph } = (/) <-> A. x -. ph )

Proof

Step Hyp Ref Expression
1 nfab1
 |-  F/_ x { x | ph }
2 1 eq0f
 |-  ( { x | ph } = (/) <-> A. x -. x e. { x | ph } )
3 abid
 |-  ( x e. { x | ph } <-> ph )
4 3 notbii
 |-  ( -. x e. { x | ph } <-> -. ph )
5 4 albii
 |-  ( A. x -. x e. { x | ph } <-> A. x -. ph )
6 2 5 bitri
 |-  ( { x | ph } = (/) <-> A. x -. ph )