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Theorem elpr2 4575
 Description: A member of a pair of sets is one or the other of them, and conversely. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.) (Proof shortened by JJ, 23-Jul-2021.)
Hypotheses
Ref Expression
elpr2.1 𝐵 ∈ V
elpr2.2 𝐶 ∈ V
Assertion
Ref Expression
elpr2 (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵𝐴 = 𝐶))

Proof of Theorem elpr2
StepHypRef Expression
1 elpr2.1 . 2 𝐵 ∈ V
2 elpr2.2 . 2 𝐶 ∈ V
3 elpr2g 4574 . 2 ((𝐵 ∈ V ∧ 𝐶 ∈ V) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵𝐴 = 𝐶)))
41, 2, 3mp2an 691 1 (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵𝐴 = 𝐶))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 209   ∨ wo 844   = wceq 1538   ∈ wcel 2115  Vcvv 3480  {cpr 4552 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-v 3482  df-un 3924  df-sn 4551  df-pr 4553 This theorem is referenced by:  elopg  5345  elxr  12508  fprodex01  30552  nofv  33221
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