Theorem elsnc2 1832

Description: There is only one element in a singleton. Exercise 2 of [TakeutiZaring] p. 15. This variation requires only that B, rather than A, be a set.

Hypothesis

Ref	Expression
elsnc2.1	⊢ B ∈ V

Assertion

Ref	Expression
elsnc2	⊢ (A ∈ {B} ↔ A = B)

Proof of Theorem elsnc2

Step	Hyp	Ref	Expression
1		elsnc2.1	. 2 ⊢ B ∈ V
2		elsnc2g 1831	. 2 ⊢ (B ∈ 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:  el1o 3115  elni 3798  elnn0 4536  seqlem2 4663  elch0 5158

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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