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Metamath Proof Explorer

Theorem nel0

Description: From the general negation of membership in A , infer that A is the empty set. (Contributed by BJ, 6-Oct-2018)

Ref Expression
Hypothesis nel0.1 
|- -. x e. A
Assertion nel0 
|- A = (/)

Proof

Step Hyp Ref Expression
1 nel0.1 
 |-  -. x e. A
2 eq0 
 |-  ( A = (/) <-> A. x -. x e. A )
3 2 1 mpgbir 
 |-  A = (/)