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Theorem nelpr 32556
Description: A set 𝐴 not in a pair is neither element of the pair. (Contributed by Thierry Arnoux, 20-Nov-2023.)
Assertion
Ref Expression
nelpr (𝐴𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴𝐵𝐴𝐶)))

Proof of Theorem nelpr
StepHypRef Expression
1 elprg 4652 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵𝐴 = 𝐶)))
21notbid 318 . 2 (𝐴𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ ¬ (𝐴 = 𝐵𝐴 = 𝐶)))
3 neanior 3032 . 2 ((𝐴𝐵𝐴𝐶) ↔ ¬ (𝐴 = 𝐵𝐴 = 𝐶))
42, 3bitr4di 289 1 (𝐴𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴𝐵𝐴𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  wo 847   = wceq 1536  wcel 2105  wne 2937  {cpr 4632
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-ext 2705
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1539  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-ne 2938  df-v 3479  df-un 3967  df-sn 4631  df-pr 4633
This theorem is referenced by:  inpr0  32557  xnn01gt  32780
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