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The completeness theorem fails for second-order logic. This question has some nice examples of consistent second-order theories without models. But non of them is finitely axiomatizable, at least those examples use infinitely many axioms.

Are there consistent finitely axiomatizable second-order theories without models, or is it possible to prove a completeness theorem for these theories?

M. Winter
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    Very interesting question. But I wonder, how would one prove the consistency of an example ? What I mean is that in the examples given in the link you provided, to prove the consistency of the theory in question, the person who answered uses the fact that any finite subtheory has a model and therefore is consistent, and so the original (infinite) theory isn't contradictory. But if the theory is finitely axiomatizable, I wonder how one would go about proving it is consistent. – Maxime Ramzi May 12 '17 at 08:31
  • @Max Good question! Maybe it is somehow obvious or provable that the given axioms (and some rules of inference) cannot prove a sentence $\varphi\wedge\neg\varphi$. I mean, he used intuition for the finite subtheories in his answer too – M. Winter May 12 '17 at 08:54
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    When you say "consistent," what proof system do you mean? Since there is no complete sound proof system for second-order logic, the notion of "consistent" is ambiguous (while it is unambiguous for first-order logic). – Noah Schweber May 12 '17 at 10:39
  • @NoahSchweber I asked this to myself. What do you think is the prove system used in the linked answer? Or is this also kind of informal over there? – M. Winter May 12 '17 at 10:40
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    @M.Winter The answers which are finitely satisfiable dodge the question - since they are finitely satisfiable, any sound proof system can't find an inconsistency (since any proof can only refer to finitely many axioms). But if you want a finite counterexample, then you do need to specify what proof system you use. – Noah Schweber May 12 '17 at 10:40

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There is an ambiguity here which arises because of your focus on finite counterexamples - what proof system are you using?

A proof system for a logic is generally taken to be at the very least a relation between sets of sentences and individual sentences which is monotonic (more axioms prove more theorems) and finitely based (if $T$ proves $\varphi$, then some finite subset of $T$ proves $\varphi$). This last condition seems out of place in the context of second-order logic, where compactness fails wildly, but captures the idea that proofs are finite objects. This rules out the possibility of complete proof systems for incompact logics, leading to examples as in the linked question, but leaves open the possibility of proof systems which are complete when restricted to finite sets of sentences.

Now the barrier becomes comparability theoretic - we probably want the proof relation to be c.e., but this prevents completeness even for finite sets of sentences in second order logic. So regardless of what proof system you use, there will be consistent finite unsatisfiable sets of sentences - but to give a concrete example we would need to specify a proof system first.

Noah Schweber
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  • Asaf constructed an example for any reasonable deductive system (compact + sound). How can I see, that such a general approach is not possible here? Can it be that there is an example with finitely many axioms that works for any compact and sound deductive system? – M. Winter May 12 '17 at 16:07
  • What is a c.e. proof relation ? – Maxime Ramzi May 13 '17 at 09:55
  • @Max I would go for something like "computably enumerable", but not sure. – M. Winter May 13 '17 at 21:14
  • @Max Yes, "c.e." means "computably enumerable" - that is, the relation "$\Gamma$ proves $\varphi$" (for $\Gamma$ a finite set of sentences and $\varphi$ a single sentence) should be computably enumerable (remember that we can code individual sentences by natural numbers in a reasonable way, so by extension we can code pairs $(\Gamma,\varphi)$ by natural numbers in a reasonable way, so this makes sense). The point is that (i) each proof should be represented by a natural number, and (ii) telling whether $n$ is a code for a proof of $\varphi$ from $\Gamma$ should be decidable. – Noah Schweber May 14 '17 at 01:27
  • @M.Winter No, there can't be a finite example that works for all compact sound deductive systems - for any unsatisfiable finite set $\Gamma$, consider the proof system that deduces $\varphi$ from $\Gamma$ for every $\varphi$. Then according to this proof system, $\Gamma$ is inconsistent; and this proof system is clearly sound (and c.e. as well). – Noah Schweber May 14 '17 at 01:28
  • @Noah Can you describe in a short form, why your first sentence does not apply to infinite axiom systems? Why is the ambiguity only there in the finite case. I got that this is how it is, but is there a short reason why? – M. Winter May 14 '17 at 16:56
  • @M.Winter Precisely because in the infinite case there is a single example which defeats all possible proof systems. It's not that the ambiguity doesn't exist, but just that it's not relevant to the particular question being asked. It would be highly relevant in different questions - e.g. "Does $\Gamma$ prove $\varphi$?" is completely dependent on the proof system used regardless of whether $\Gamma$ is finite. But since in this case we have a negative result that works for all proof systems, the choice of proof system doesn't really come up in the same way. – Noah Schweber May 14 '17 at 17:19