Theorem elprn2 4602

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 4597	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2	1	adantr 480	. 2 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
3		neneq 2934	. . 3 ⊢ (𝐴 ≠ 𝐶 → ¬ 𝐴 = 𝐶)
4	3	adantl 481	. 2 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → ¬ 𝐴 = 𝐶)
5	2, 4	olcnd 877	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐴 ≠ 𝐶) → 𝐴 = 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∨ wo 847   = wceq 1541   ∈ wcel 2111   ≠ wne 2928  {cpr 4575

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-v 3438  df-un 3902  df-sn 4574  df-pr 4576

This theorem is referenced by:  esplymhp  33589