Theorem elprneb 47826

Description: An element of a proper unordered pair is the first element iff it is not the second element. (Contributed by AV, 18-Jun-2020.)

Assertion

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

Proof of Theorem elprneb

Step	Hyp	Ref	Expression
1		elpri 4615	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2		neeq1 3022	. . . . . 6 ⊢ (𝐵 = 𝐴 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
3	2	eqcoms 2773	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
4		pm5.1 836	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
5	4	ex 418	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
6	3, 5	sylbid 243	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
7		neeq2 3023	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 ↔ 𝐵 ≠ 𝐶))
8		nesym 3016	. . . . . . . 8 ⊢ (𝐵 ≠ 𝐴 ↔ ¬ 𝐴 = 𝐵)
9		pm5.1 836	. . . . . . . 8 ⊢ ((𝐴 = 𝐶 ∧ ¬ 𝐴 = 𝐵) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
10	8, 9	sylan2b 606	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
11	10	necon2abid 3002	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
12	11	ex 418	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
13	7, 12	sylbird 263	. . . 4 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
14	6, 13	jaoi 871	. . 3 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
15	1, 14	syl 18	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
16	15	imp 412	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 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 2146   ≠ wne 2960  {cpr 4593

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 2148  ax-9 2156  ax-ext 2737

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 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-v 3459  df-un 3911  df-sn 4592  df-pr 4594

This theorem is used by:  dfodd5  48485