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SOP_2 = SOP_3

Logic & foundations · posed by Mirna Džamonja, Saharon Shelah, 2004 · solved

1 comment

Statement

The classes of SOP_2 and SOP_3 first-order theories coincide. This answers a question of Džamonja and Shelah from 2004.

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No person has examined this. Nothing here has been checked at all. say whether it holds →

  1. proof attempt · #1

    Artem Chernikov, using ChatGPT 5.6

    That credit came with the record as it was imported. No ProbXiv account is credited for this work, and nobody has answered for it here.

    AI involvement
    ai co developed
    — a person and a model developed the result together.

    From the paper: "The proof was found using ChatGPT 5.6 and simplified and streamlined by the author."

    The new content is SOP2⇒SOP3SOP_2 \Rightarrow SOP_3; the converse implication was known from the start. Dzamonja and Shelah asked whether either implication in SOP3⇒SOP2⇒SOP1SOP_3 \Rightarrow SOP_2 \Rightarrow SOP_1 reverses: Mutchnik answered the second (SOP1=SOP2SOP_1 = SOP_2), and this answers the first, collapsing the bottom of the hierarchy to SOP1=SOP2=SOP3SOP_1 = SOP_2 = SOP_3. The SOPnSOP_n hierarchy for n≥3n \ge 3 remains, as does everything above it.

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