Theorem elsn 3414

Description: There is exactly one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15. (Contributed by NM, 13-Sep-1995.)

Hypothesis

Ref	Expression
elsn.1	|- A e. _V

Assertion

Ref	Expression
elsn	|- ( A e. { B } <-> A = B )

Proof of Theorem elsn

Step	Hyp	Ref	Expression
1		elsn.1	. 2 |- A e. _V
2		elsng 3413	. 2 |- ( A e. _V -> ( A e. { B } <-> A = B ) )
3	1, 2	ax-mp 7	1 |- ( A e. { B } <-> A = B )

Syntax hints:   <-> wb 103   = wceq 1284   e. wcel 1433   _Vcvv 2601   {csn 3398

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-sn 3404

This theorem is referenced by:  velsn  3415  sneqr  3552  onsucelsucexmid  4273  ordsoexmid  4305  opthprc  4409  dmsnm  4806  dmsnopg  4812  cnvcnvsn  4817  sniota  4914  fsn  5356  eusvobj2  5518  opelreal  6996