Theorem nel0 3430

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

Hypothesis

Ref	Expression
nel0.1	|- -. x e. A

Assertion

Ref	Expression
nel0	|- A = (/)

Distinct variable group:   x, A

Proof of Theorem nel0

Step	Hyp	Ref	Expression
1		eq0 3427	. 2 |- ( A = (/) <-> A. x -. x e. A )
2		nel0.1	. 2 |- -. x e. A
3	1, 2	mpgbir 1441	1 |- A = (/)

Syntax hints:   -. wn 3   = wceq 1343   e. wcel 2136   (/)c0 3409

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-dif 3118  df-nul 3410

This theorem is referenced by: (None)