Theorem elop 1894

Description: An ordered pair has two elements. Exercise 3 of [TakeutiZaring] p. 15.

Hypothesis

Ref	Expression
elop.1	⊢ A ∈ V

Assertion

Ref	Expression
elop	⊢ (A ∈ ⟨B, C⟩ ↔ (A = {B} ∨ A = {B, C}))

Proof of Theorem elop

Step	Hyp	Ref	Expression
1		df-op 1815	. . 3 ⊢ ⟨B, C⟩ = {{B}, {B, C}}
2	1	eleq2i 1153	. 2 ⊢ (A ∈ ⟨B, C⟩ ↔ A ∈ {{B}, {B, C}})
3		elop.1	. . 3 ⊢ A ∈ V
4	3	elpr 1823	. 2 ⊢ (A ∈ {{B}, {B, C}} ↔ (A = {B} ∨ A = {B, C}))
5	2, 4	bitr 151	1 ⊢ (A ∈ ⟨B, C⟩ ↔ (A = {B} ∨ A = {B, C}))

Syntax hints:   ↔ wb 127   ∨ wo 195   = wceq 1091   ∈ wcel 1092  Vcvv 1348  {csn 1808  {cpr 1809  ⟨cop 1810

This theorem is referenced by:  opth 1898  opprc1b 1906

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  df-op 1815

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