Theorem elsncg 1825

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

Assertion

Ref	Expression
elsncg	⊢ (A ∈ C → (A ∈ {B} ↔ A = B))

Proof of Theorem elsncg

Step	Hyp	Ref	Expression
1		elprg 1822	. 2 ⊢ (A ∈ C → (A ∈ {B, B} ↔ (A = B ∨ A = B)))
2		dfsn2 1819	. . . 4 ⊢ {B} = {B, B}
3	2	cleqcomi 1105	. . 3 ⊢ {B, B} = {B}
4	3	eleq2i 1153	. 2 ⊢ (A ∈ {B, B} ↔ A ∈ {B})
5		oridm 208	. 2 ⊢ ((A = B ∨ A = B) ↔ A = B)
6	1, 4, 5	3bitr3g 427	1 ⊢ (A ∈ C → (A ∈ {B} ↔ A = B))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   = wceq 1091   ∈ wcel 1092  {csn 1808  {cpr 1809

This theorem is referenced by:  elsnc 1826  elsni 1827  snidg 1828  elsucg 2290

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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