Theorem elsn 3422

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 3421	. 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 1285   e. wcel 1434   _Vcvv 2602   {csn 3406

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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-v 2604  df-sn 3412

This theorem is referenced by:  velsn  3423  sneqr  3560  onsucelsucexmid  4281  ordsoexmid  4313  opthprc  4417  dmsnm  4816  dmsnopg  4822  cnvcnvsn  4827  sniota  4924  fsn  5367  eusvobj2  5529  opelreal  7058