Theorem nelpr 33061

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 4606	. . 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 3048	. 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 401   ∨ wo 861   = wceq 1570   ∈ wcel 2145   ≠ wne 2955  {cpr 4585

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-v 3452  df-un 3903  df-sn 4584  df-pr 4586

This theorem is used by:  inpr0  33062  xnn01gt  33296