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Theorem nelpr 30879
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 4582 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵𝐴 = 𝐶)))
21notbid 318 . 2 (𝐴𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ ¬ (𝐴 = 𝐵𝐴 = 𝐶)))
3 neanior 3037 . 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 205  wa 396  wo 844   = wceq 1539  wcel 2106  wne 2943  {cpr 4563
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ne 2944  df-v 3434  df-un 3892  df-sn 4562  df-pr 4564
This theorem is referenced by:  inpr0  30880  xnn01gt  31093
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