Theorem elprneb 42102

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 4420	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2		neeq1 3031	. . . . . 6 ⊢ (𝐵 = 𝐴 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
3	2	eqcoms 2786	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
4		pm5.1 815	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
5	4	ex 403	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
6	3, 5	sylbid 232	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
7		neeq2 3032	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 ↔ 𝐵 ≠ 𝐶))
8		nesym 3025	. . . . . . . 8 ⊢ (𝐵 ≠ 𝐴 ↔ ¬ 𝐴 = 𝐵)
9		pm5.1 815	. . . . . . . 8 ⊢ ((𝐴 = 𝐶 ∧ ¬ 𝐴 = 𝐵) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
10	8, 9	sylan2b 587	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
11	10	necon2abid 3011	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
12	11	ex 403	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
13	7, 12	sylbird 252	. . . 4 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
14	6, 13	jaoi 846	. . 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 397	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐵 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 198   ∧ wa 386   ∨ wo 836   = wceq 1601   ∈ wcel 2107   ≠ wne 2969  {cpr 4400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-ext 2754

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-ne 2970  df-v 3400  df-un 3797  df-sn 4399  df-pr 4401

This theorem is referenced by:  dfodd5  42601