Theorem sotritric 4204

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 4203	. . 3 ⊢ ((𝑅 Or 𝐴 ∧ (𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴)) → (𝐵𝑅𝐶 → ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
3	1, 2	mpan 418	. 2 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 → ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
4		sotritric.tri	. . 3 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵))
5		3orass 946	. . . 4 ⊢ ((𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) ↔ (𝐵𝑅𝐶 ∨ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))
6		ax-1 5	. . . . 5 ⊢ (𝐵𝑅𝐶 → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
7		pm2.24 593	. . . . 5 ⊢ ((𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
8	6, 7	jaoi 688	. . . 4 ⊢ ((𝐵𝑅𝐶 ∨ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)) → (¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵) → 𝐵𝑅𝐶))
9	5, 8	sylbi 120	. . 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 128	1 ⊢ ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → (𝐵𝑅𝐶 ↔ ¬ (𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 680   ∨ w3o 942   = wceq 1312   ∈ wcel 1461   class class class wbr 3893   Or wor 4175

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 586  ax-in2 587  ax-io 681  ax-5 1404  ax-7 1405  ax-gen 1406  ax-ie1 1450  ax-ie2 1451  ax-8 1463  ax-10 1464  ax-11 1465  ax-i12 1466  ax-bndl 1467  ax-4 1468  ax-17 1487  ax-i9 1491  ax-ial 1495  ax-i5r 1496  ax-ext 2095

This theorem depends on definitions:  df-bi 116  df-3or 944  df-3an 945  df-tru 1315  df-nf 1418  df-sb 1717  df-clab 2100  df-cleq 2106  df-clel 2109  df-nfc 2242  df-ral 2393  df-v 2657  df-un 3039  df-sn 3497  df-pr 3498  df-op 3500  df-br 3894  df-po 4176  df-iso 4177

This theorem is referenced by:  nqtric  7148