Theorem sotritric 4415

Description: A trichotomy relationship, given a trichotomous order. (Contributed by Jim Kingdon, 28-Sep-2019.)

Hypotheses

Ref	Expression
sotritric.or	⊢ 𝑅 Or 𝐴
sotritric.tri	⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵))

Assertion

Ref	Expression
sotritric	⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ↔ ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))

Proof of Theorem sotritric

Step	Hyp	Ref	Expression
1		sotritric.or	. . 3 ⊢ 𝑅 Or 𝐴
2		sotricim 4414	. . 3 ⊢ ((𝑅 Or 𝐴 ∧ (𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴)) → (𝐵𝑅𝐶 → ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
3	1, 2	mpan 424	. 2 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 → ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
4		sotritric.tri	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵))
5		3orass 1005	. . . 4 ⊢ ((𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) ↔ (𝐵𝑅𝐶 ∨ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
6		ax-1 6	. . . . 5 ⊢ (𝐵𝑅𝐶 → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
7		pm2.24 624	. . . . 5 ⊢ ((𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
8	6, 7	jaoi 721	. . . 4 ⊢ ((𝐵𝑅𝐶 ∨ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
9	5, 8	sylbi 121	. . 3 ⊢ ((𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
10	4, 9	syl 14	. 2 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
11	3, 10	impbid 129	1 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ↔ ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 713   ∨ w3o 1001   = wceq 1395   ∈ wcel 2200   class class class wbr 4083   Or wor 4386

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-3or 1003  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-v 2801  df-un 3201  df-sn 3672  df-pr 3673  df-op 3675  df-br 4084  df-po 4387  df-iso 4388

This theorem is referenced by:  nqtric  7586