Every first-order theory with the tree property SOP2 also has SOP3, so SOP1, SOP2, and SOP3 define one identical dividing line.
Independence over arbitrary sets inNSOP1 theories.Annals of Pure and Applied Logic, 173(2):103058, 2022
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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.