Theorem elsnc 1826

Description: There is only one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15.

Hypothesis

Ref	Expression
elsnc.1	⊢ A ∈ V

Assertion

Ref	Expression
elsnc	⊢ (A ∈ {B} ↔ A = B)

Proof of Theorem elsnc

Step	Hyp	Ref	Expression
1		elsnc.1	. 2 ⊢ A ∈ V
2		elsncg 1825	. 2 ⊢ (A ∈ V → (A ∈ {B} ↔ A = B))
3	1, 2	ax-mp 6	1 ⊢ (A ∈ {B} ↔ A = B)

Syntax hints:   ↔ wb 127   = wceq 1091   ∈ wcel 1092  Vcvv 1348  {csn 1808

This theorem is referenced by:  eltp 1834  sneqr 1856  opth 1898  opthwiener 1914  snsn0non 2371  opthprc 2457  dmsn0 2543  dmsnsn0 2544  dmsnop 2547  cnvsn 2636  funsn 2690  fsn 2895  1st2val 3097  limenpsi 3400  opelreal 4043  divval 4217  ruclem8 4892  hsn0elch 5155  h1de2ctlem 5460

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-sn 1811  df-pr 1812

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