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Currently, the most common definition of set theory is built from first-order logic, which means that to include substitution requires an infinite schema of axioms for every possible case because of first-order logic's limitations to scope of variables.

Would basing set theory in second-order logic mean we can rewrite the schema of substitution as a single axiom? Are there practical benefits to this? Are they worth losing completeness and compactness from first-order logic, or are they equivalent (as in we just switch the 'problem' from sets to the language)?

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    As far as concerns what can be proved, this is essentially what MK set theory does, except that in most presentation it disguises the second-order variables as a separate kind of first-order obejcts. You then still need an axiom schema to guarantee the existence of sufficiently many classes, though -- so not much is gained in terms of foundational complexity. – hmakholm left over Monica Jul 12 '17 at 17:07
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    @Henning: Actually, not quite. Second order ZF is equivalent to inaccessible cardinals; MK does not imply the consistency of inaccessible cardinals, though. – Asaf Karagila Jul 12 '17 at 17:10
  • @Henning: Also note that MK has a Replacement schema since it allows impredicative quantification over second-order variables in the schema. So turning this into a single axiom would require third-order logic. – Asaf Karagila Jul 12 '17 at 17:13
  • @AsafKaragila: That may be ... I'm not exactly sure about which reasoning principles for proving predicate existence are usually implied by saying "second-order logic". I sort of assumed it would allow impredicative ones, since that seems to be sound with respect to the standard semantics. – hmakholm left over Monica Jul 12 '17 at 17:18
  • @industryfatcat I think it's a good idea, I'm researching it myself. It seems intuitive. Asaf pointed out some disadvantages, but maybe the advantages are greater. – Paul Feb 20 '18 at 23:52

1 Answers1

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The thing about the second-order of $\sf ZFC$ is that $M$ is a model of second-order $\sf ZFC$ if and only if $M$ is isomorphic to $V_\kappa$ for an inaccessible $\kappa$. So right off the bat you start with a much stronger consistency assumptions. Okay, that might not be a real problem. I mean, we believe $\sf ZFC$ is consistent, what's one more inaccessible cardinal, eh? But wait, there's more!

Not only that we lose a bunch of useful tools from first-order logic, we also seemingly determine things like the truth value of $\sf CH$ (since every model must contain all the reals and sets of reals, so they must agree on the truth value of $\sf CH$), and so we lose the ability to use forcing as a tool for proving certain statements.

Moreover, there is an inherent problem basing set theory on second-order logic: the notion of a set is more or less required for the development of second-order logic, so your meta-theory should already include some version of set theory to begin with.

Now, normally this is not a terrible thing where your meta-theory is the same as your theory (in principle, anyway), since we often treat the meta-theory of $\sf ZFC$ as $\sf ZFC$ itself again. But therein lies the rub: first-order is nice and recursive and we can replace the meta-theory from $\sf ZFC$ to something stupendously weak like $\sf PRA$, and all the proofs remain more or less the same (except schemata of proofs done in the meta-theory which are now slightly more complicated); whereas with second-order logic you cannot really weaken your theory too much.

And even if you could, you lose the whole point for which we fight for first-order logic. You lose the ability to recursively identify and check the correctness of proofs. This is one of the good reasons that a set theoretic foundation is useful. Since second-order logic does not have a nicely algorithmic proof theory to it, moving from second-order things (like analysis and topology which are even of higher-order) to first-order statements like "$\sf ZFC$ proves bla bla bla" allows us to compensate for this problem.

On top of all this, second-order $\sf ZFC$ can have multiple models and it is not complete in any sense of the word (for example, if $\kappa<\lambda$ are two inaccessible cardinals, then $V_\kappa$ and $V_\lambda$ are both models of second-order $\sf ZFC$, but only one of them satisfies "There exists an inaccessible cardinal").

To sum up, here are the main problems:

  1. Second-order $\sf ZFC$ requires a stronger consistency assumptions to produce models of the theory.
  2. Second-order logic requires a notion of "set" from the get-go, so some set theoretic meta-theory is needed.
  3. Second-order $\sf ZFC$ forces us to accept whatever the meta-theory decided on a variety of statements like $\sf CH$, making the role of the meta-theory much greater than we would have liked to in our mathematics.
  4. Second-order logic has a lousy proof theory to it, so you lose all that you wanted to gain in terms of automatic proof verification for "everyday mathematics" which is often done in high-order logic and using $\sf ZFC$ we translate it back to first-order logic.
Asaf Karagila
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    What is the distinction in #4 between high-order logic used in verification for everyday mathematics (which can just be translated to FOL) versus second-order logic, which you say has a lousy proof theory. Isn't second-order logic just a special case of higher-order logic, and can also be translated to ZFC for proof verification purposes? (I assume I'm just missing something obvious) – Andy Nov 03 '20 at 23:04
  • Yes, you are missing something. The goal here is to consider a foundation of mathematics in second-order logic, not formalise it in ZFC... :-) – Asaf Karagila Nov 03 '20 at 23:25
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    Thanks. When you say 2nd-order has lousy proof theory is that just referring to incompleteness of the deductive systems, or are they are inefficient in some more practical sense, or is it some other issue? Maybe this should be separate question? – Andy Nov 04 '20 at 00:06
  • @Andy: The former. Indeed, it is the lack of a proof verification algorithm that means that there really isn't much proof theory in a practical sense. Last year I was in a conference mostly populated by type theorist, and one of them told me that he knows his system is not decidable, but he doesn't care because he lets the user input data as an oracle. Unfortunately, proof verification in mathematics is exactly meant to exclude "user input data" beyond the basic axioms. – Asaf Karagila Nov 04 '20 at 00:08