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Feb 26, 2013 at 19:55 vote accept Samuel Reid
Feb 4, 2013 at 20:04 history edited Noah Schweber CC BY-SA 3.0
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Feb 4, 2013 at 20:00 comment added Noah Schweber @Emil: thanks for the citation! I spent a while looking for that paper with no luck. Re: your first point on terminology, I recall learning the opposite convention, that $R$ being well-founded meant that every $R$-increasing sequence terminates, but looking through my notes/books I can't see where I got that convention; so maybe I'm just conflating the picture with that of descriptive set theory (where, if your trees grow upwards, a tree is well-founded if every increasing chain terminates).
Feb 4, 2013 at 13:56 comment added Emil Jeřábek It’s S. K. Thomason, Reduction of second-order logic to modal logic, Zeitschrift für mathematische Logik und Grundlagen der Mathematik 21 (1975), no. 1, 107–114. One can in fact fix $\phi$, so the reduction is to Kripke validity in a particular finitely axiomatized modal logic. (Thomason only states it for monadic second-order logic, but as far as I can see, this makes no difference, as one can encode a pairing function in the binary relation.)
Feb 4, 2013 at 13:02 comment added Emil Jeřábek ... is recursively reducible to the relation “$\psi$ is valid in every Kripke frame in which $\phi$ is valid”.
Feb 4, 2013 at 13:01 comment added Emil Jeřábek Well, $R\subseteq X^2$ is well-founded if every nonempty subset of $X$ has an $R$-minimal element, and converse well-founded if every such set has an $R$-maximal element. These are distinct properties, and GL corresponds to transitive frames with the latter property. As for the connection to second-order logic: modally definable second-order properties are indeed rather special (e.g., they are monadic $\Pi^1_1$, and they are preserved by the four operations listed), but the “sort of converse” you mention might be the result (due to Thomason, IIRC) that validity in full second-order logic ...
Feb 3, 2013 at 23:52 history edited Noah Schweber CC BY-SA 3.0
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Feb 3, 2013 at 20:19 history answered Noah Schweber CC BY-SA 3.0