Theorem nel0 3415

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 3412	. 2 |- ( A = (/) <-> A. x -. x e. A )
2		nel0.1	. 2 |- -. x e. A
3	1, 2	mpgbir 1433	1 |- A = (/)

Syntax hints:   -. wn 3   = wceq 1335   e. wcel 2128   (/)c0 3394

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 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139

This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-v 2714  df-dif 3104  df-nul 3395

This theorem is referenced by: (None)