Theorem elprn2 4612

Description: A member of an unordered pair that is not the "second", must be the "first". (Contributed by Glauco Siliprandi, 11-Dec-2019.)

Assertion

Ref	Expression
elprn2	⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → 𝐴 = 𝐵)

Proof of Theorem elprn2

Step	Hyp	Ref	Expression
1		elpri 4607	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2	1	adantr 486	. 2 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
3		neneq 2964	. . 3 ⊢ (𝐴 ≠ 𝐶 → ¬ 𝐴 = 𝐶)
4	3	adantl 487	. 2 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → ¬ 𝐴 = 𝐶)
5	2, 4	olcnd 891	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → 𝐴 = 𝐵)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1569   ∈ wcel 2146   ≠ wne 2958  {cpr 4589

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-v 3455  df-un 4153  df-sn 4588  df-pr 4590

This theorem is used by:  esplymhp  34253