Every first-order theory with the tree property SOP2 also has SOP3, so SOP1, SOP2, and SOP3 define one identical dividing line.
Transitivity, lowness, and ranks inNSOP1 theories.The Journal of Symbolic Logic, 88(3):919–946, 2023
1 Pith paper cite this work. Polarity classification is still indexing.
1
Pith paper citing it
fields
math.LO 1years
2026 1verdicts
CONDITIONAL 1representative citing papers
citing papers explorer
-
SOP$_2$=SOP$_3$
Every first-order theory with the tree property SOP2 also has SOP3, so SOP1, SOP2, and SOP3 define one identical dividing line.