Theorem nelpr 32609

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	⊢ (𝐴 ∈ 𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶)))

Proof of Theorem nelpr

Step	Hyp	Ref	Expression
1		elprg 4604	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
2	1	notbid 318	. 2 ⊢ (𝐴 ∈ 𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ ¬ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
3		neanior 3026	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶) ↔ ¬ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
4	2, 3	bitr4di 289	1 ⊢ (𝐴 ∈ 𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 848   = wceq 1542   ∈ wcel 2114   ≠ wne 2933  {cpr 4583

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 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-v 3443  df-un 3907  df-sn 4582  df-pr 4584

This theorem is referenced by:  inpr0  32610  xnn01gt  32852