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Theorem elop 3968
 Description: An ordered pair has two elements. Exercise 3 of [TakeutiZaring] p. 15. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 26-Apr-2015.)
Hypotheses
Ref Expression
elop.1 𝐴 ∈ V
elop.2 𝐵 ∈ V
elop.3 𝐶 ∈ V
Assertion
Ref Expression
elop (𝐴 ∈ ⟨𝐵, 𝐶⟩ ↔ (𝐴 = {𝐵} ∨ 𝐴 = {𝐵, 𝐶}))

Proof of Theorem elop
StepHypRef Expression
1 elop.2 . . . 4 𝐵 ∈ V
2 elop.3 . . . 4 𝐶 ∈ V
31, 2dfop 3548 . . 3 𝐵, 𝐶⟩ = {{𝐵}, {𝐵, 𝐶}}
43eleq2i 2104 . 2 (𝐴 ∈ ⟨𝐵, 𝐶⟩ ↔ 𝐴 ∈ {{𝐵}, {𝐵, 𝐶}})
5 elop.1 . . 3 𝐴 ∈ V
65elpr 3396 . 2 (𝐴 ∈ {{𝐵}, {𝐵, 𝐶}} ↔ (𝐴 = {𝐵} ∨ 𝐴 = {𝐵, 𝐶}))
74, 6bitri 173 1 (𝐴 ∈ ⟨𝐵, 𝐶⟩ ↔ (𝐴 = {𝐵} ∨ 𝐴 = {𝐵, 𝐶}))
 Colors of variables: wff set class Syntax hints:   ↔ wb 98   ∨ wo 629   = wceq 1243   ∈ wcel 1393  Vcvv 2557  {csn 3375  {cpr 3376  ⟨cop 3378 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-un 2922  df-sn 3381  df-pr 3382  df-op 3384 This theorem is referenced by:  relop  4486  bdop  9995
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