Theorem sotrieq2 2150

Description: Trichotomy law for strict order relation.

Assertion

Ref	Expression
sotrieq2	⊢ ((R Or A ∧ (B ∈ A ∧ C ∈ A)) → (B = C ↔ (¬ BRC ∧ ¬ CRB)))

Proof of Theorem sotrieq2

Step	Hyp	Ref	Expression
1		sotrieq 2149	. 2 ⊢ ((R Or A ∧ (B ∈ A ∧ C ∈ A)) → (B = C ↔ ¬ (BRC ∨ CRB)))
2		ioran 254	. 2 ⊢ (¬ (BRC ∨ CRB) ↔ (¬ BRC ∧ ¬ CRB))
3	1, 2	syl6bb 414	1 ⊢ ((R Or A ∧ (B ∈ A ∧ C ∈ A)) → (B = C ↔ (¬ BRC ∧ ¬ CRB)))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∨ wo 195   ∧ wa 196   = wceq 1091   ∈ wcel 1092   class class class wbr 2054   Or wor 2059

This theorem is referenced by:  supmo 2156  lttri3t 4281

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-3an 583  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-v 1349  df-un 1490  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-po 2128  df-so 2138

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