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Theorem brtp 5498
Description: A necessary and sufficient condition for two sets to be related under a binary relation which is an unordered triple. (Contributed by Scott Fenton, 8-Jun-2011.)
Hypotheses
Ref Expression
brtp.1 𝑋 ∈ V
brtp.2 𝑌 ∈ V
Assertion
Ref Expression
brtp (𝑋{⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩}𝑌 ↔ ((𝑋 = 𝐴𝑌 = 𝐵) ∨ (𝑋 = 𝐶𝑌 = 𝐷) ∨ (𝑋 = 𝐸𝑌 = 𝐹)))

Proof of Theorem brtp
StepHypRef Expression
1 df-br 5120 . 2 (𝑋{⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩}𝑌 ↔ ⟨𝑋, 𝑌⟩ ∈ {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩})
2 opex 5439 . . 3 𝑋, 𝑌⟩ ∈ V
32eltp 4665 . 2 (⟨𝑋, 𝑌⟩ ∈ {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩} ↔ (⟨𝑋, 𝑌⟩ = ⟨𝐴, 𝐵⟩ ∨ ⟨𝑋, 𝑌⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝑋, 𝑌⟩ = ⟨𝐸, 𝐹⟩))
4 brtp.1 . . . 4 𝑋 ∈ V
5 brtp.2 . . . 4 𝑌 ∈ V
64, 5opth 5451 . . 3 (⟨𝑋, 𝑌⟩ = ⟨𝐴, 𝐵⟩ ↔ (𝑋 = 𝐴𝑌 = 𝐵))
74, 5opth 5451 . . 3 (⟨𝑋, 𝑌⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝑋 = 𝐶𝑌 = 𝐷))
84, 5opth 5451 . . 3 (⟨𝑋, 𝑌⟩ = ⟨𝐸, 𝐹⟩ ↔ (𝑋 = 𝐸𝑌 = 𝐹))
96, 7, 83orbi123i 1156 . 2 ((⟨𝑋, 𝑌⟩ = ⟨𝐴, 𝐵⟩ ∨ ⟨𝑋, 𝑌⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝑋, 𝑌⟩ = ⟨𝐸, 𝐹⟩) ↔ ((𝑋 = 𝐴𝑌 = 𝐵) ∨ (𝑋 = 𝐶𝑌 = 𝐷) ∨ (𝑋 = 𝐸𝑌 = 𝐹)))
101, 3, 93bitri 297 1 (𝑋{⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩}𝑌 ↔ ((𝑋 = 𝐴𝑌 = 𝐵) ∨ (𝑋 = 𝐶𝑌 = 𝐷) ∨ (𝑋 = 𝐸𝑌 = 𝐹)))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395  w3o 1085   = wceq 1540  wcel 2108  Vcvv 3459  {ctp 4605  cop 4607   class class class wbr 5119
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707  ax-sep 5266  ax-nul 5276  ax-pr 5402
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3or 1087  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-rab 3416  df-v 3461  df-dif 3929  df-un 3931  df-ss 3943  df-nul 4309  df-if 4501  df-sn 4602  df-pr 4604  df-tp 4606  df-op 4608  df-br 5120
This theorem is referenced by:  sltval2  27620  sltintdifex  27625  sltres  27626  noextendlt  27633  noextendgt  27634  nolesgn2o  27635  nogesgn1o  27637  sltsolem1  27639  nosepnelem  27643  nosep1o  27645  nosep2o  27646  nosepdmlem  27647  nodenselem8  27655  nodense  27656  nolt02o  27659  nogt01o  27660  nosupbnd2lem1  27679  noinfbnd2lem1  27694
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