Theorem nelpr 32888

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 4611	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
2	1	notbid 321	. 2 ⊢ (𝐴 ∈ 𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ ¬ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
3		neanior 3050	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶) ↔ ¬ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
4	2, 3	bitr4di 292	1 ⊢ (𝐴 ∈ 𝑉 → (¬ 𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1569   ∈ wcel 2142   ≠ wne 2957  {cpr 4590

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-v 3456  df-un 3909  df-sn 4589  df-pr 4591

This theorem is used by:  inpr0  32889  xnn01gt  33126