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Apr 28, 2020 at 15:58 comment added Jori @NoahSchweber I think that is a very reasonable articulation of my rather vague use of "ab initio". So then what is SOL useful for? Does it have any use in possibly deciding CH? It seems somewhat dishonest to call something a logic which has a sentence that is a logical truth iff CH is really true, but without it having any bearing on CH...
Apr 26, 2020 at 16:59 comment added Noah Schweber Basically, "understand $\mathcal{L}$ ab initio" should probably mean "the basic entailment notion of $\mathcal{L}$ shouldn't depend on set-theoretic problems" - and SOL fails this test terribly. (If that's not what you mean by that, then what do you mean?)
Apr 26, 2020 at 16:58 comment added Noah Schweber @Jori "Can't we understand SOL ab initio?" No, not really, for exactly the reasons in my answer. There is a second-order sentence $\chi$ which is a validity iff CH is true, for example. So asking whether $\chi$ is a "second-order tautology" amounts to asking whether CH is true.
Apr 26, 2020 at 16:56 comment added Noah Schweber @Jori "we need a set-theoretic metatheory in order to make sense of the semantics of SOL" Yeah, basically. In order to tell whether a second-order sentence is true in a structure $\mathcal{A}$, we need in general to have access to the whole powerset of $\mathcal{A}$ (in contrast to first-order logic). So for example in order to talk about whether a second-order sentence is true of the natural numbers, we need to admit completed infinite sets into our framework (to quantify over them).
Apr 26, 2020 at 16:48 comment added Jori What do you mean by "note that in order for second-order logic to make sense, we have to make a commitment to an underlying 'real' universe of sets"? Something like: we need a set-theoretic metatheory in order to make sense of the semantics of SOL? Is that really true? Can't we understand SOL ab initio?
Oct 21, 2017 at 21:47 comment added Mike Battaglia This answer was many years ago, but in case anyone is still reading, here is a followup. Does the same concept apply to ZF2 for AC? That is, I understand that ZFC2, with the additional axiom that there are no strongly inaccessible cardinals, is categorical, hence CH is decided one way or another (though we don't know how). Do we have the same exact situation with ZF2? Do large cardinals change the story at all?
Oct 14, 2011 at 18:32 comment added Garabed Gulbenkian Actually I thought that my statement in quotes is equivalent to your statement in quotes, but maybe I am missing something.
Oct 14, 2011 at 18:21 comment added Garabed Gulbenkian See my comment in response to the answer of Andreas above.
Oct 14, 2011 at 4:11 history edited Noah Schweber CC BY-SA 3.0
Removed redundant language
Oct 14, 2011 at 3:37 history edited Noah Schweber CC BY-SA 3.0
Forgot to actually answer the question
Oct 13, 2011 at 23:50 history answered Noah Schweber CC BY-SA 3.0