Theorem elprneb 41296

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 4197	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
2		neeq1 2856	. . . . . 6 ⊢ (𝐵 = 𝐴 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
3	2	eqcoms 2630	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 ↔ 𝐴 ≠ 𝐶))
4		pm5.1 902	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
5	4	ex 450	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
6	3, 5	sylbid 230	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
7		neeq2 2857	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 ↔ 𝐵 ≠ 𝐶))
8		nesym 2850	. . . . . . . 8 ⊢ (𝐵 ≠ 𝐴 ↔ ¬ 𝐴 = 𝐵)
9		pm5.1 902	. . . . . . . 8 ⊢ ((𝐴 = 𝐶 ∧ ¬ 𝐴 = 𝐵) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
10	8, 9	sylan2b 492	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐶 ↔ ¬ 𝐴 = 𝐵))
11	10	necon2abid 2836	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 ≠ 𝐴) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
12	11	ex 450	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐴 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
13	7, 12	sylbird 250	. . . 4 ⊢ (𝐴 = 𝐶 → (𝐵 ≠ 𝐶 → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶)))
14	6, 13	jaoi 394	. . 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 445	1 ⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐵 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∨ wo 383   ∧ wa 384   = wceq 1483   ∈ wcel 1990   ≠ wne 2794  {cpr 4179

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-v 3202  df-un 3579  df-sn 4178  df-pr 4180

This theorem is referenced by:  dfodd5  41572