Metamath Proof Explorer


Theorem eq0f

Description: A class is equal to the empty set if and only if it has no elements. Theorem 2 of Suppes p. 22. (Contributed by BJ, 15-Jul-2021)

Ref Expression
Hypothesis eq0f.1
|- F/_ x A
Assertion eq0f
|- ( A = (/) <-> A. x -. x e. A )

Proof

Step Hyp Ref Expression
1 eq0f.1
 |-  F/_ x A
2 nfcv
 |-  F/_ x (/)
3 1 2 cleqf
 |-  ( A = (/) <-> A. x ( x e. A <-> x e. (/) ) )
4 noel
 |-  -. x e. (/)
5 4 nbn
 |-  ( -. x e. A <-> ( x e. A <-> x e. (/) ) )
6 5 albii
 |-  ( A. x -. x e. A <-> A. x ( x e. A <-> x e. (/) ) )
7 3 6 bitr4i
 |-  ( A = (/) <-> A. x -. x e. A )