Theorem nelpr 32559

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 4670	. . 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 3041	. 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 846   = wceq 1537   ∈ wcel 2108   ≠ wne 2946  {cpr 4650

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947  df-v 3490  df-un 3981  df-sn 4649  df-pr 4651

This theorem is referenced by:  inpr0  32560  xnn01gt  32777