Theorem elprneb 44410

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 4580	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2		neeq1 3005	. . . . . 6 ⊢ (𝐵 = 𝐴 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
3	2	eqcoms 2746	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
4		pm5.1 820	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
5	4	ex 412	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
6	3, 5	sylbid 239	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
7		neeq2 3006	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 ↔ 𝐵 ≠ 𝐶))
8		nesym 2999	. . . . . . . 8 ⊢ (𝐵 ≠ 𝐴 ↔ ¬ 𝐴 = 𝐵)
9		pm5.1 820	. . . . . . . 8 ⊢ ((𝐴 = 𝐶 ∧ ¬ 𝐴 = 𝐵) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
10	8, 9	sylan2b 593	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
11	10	necon2abid 2985	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
12	11	ex 412	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
13	7, 12	sylbird 259	. . . 4 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
14	6, 13	jaoi 853	. . 3 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
15	1, 14	syl 17	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
16	15	imp 406	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐵 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843   = wceq 1539   ∈ wcel 2108   ≠ wne 2942  {cpr 4560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ne 2943  df-v 3424  df-un 3888  df-sn 4559  df-pr 4561

This theorem is referenced by:  dfodd5  45000