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Consider the theory $\text{Th}(\mathbb{N}, 0, S)$, with $S$ being the successor function. In his book A Course in Model Theory, Poizat claims (p. 109) that this theory is not finitely axiomatizable. Intuitively, this is because we need to employ an axiom schema such that, for each $n$, there is an axiom saying that there is no cycle of length $n$ (i.e., $S^n(x) \neq x$, where $S^n(x)$ denotes the application of $S$ to $x$ $n$ times). That is fine as far as it goes, but how would one prove that this theory is not finitely axiomatizable? I mean, it is clear that no finite subset of this axiomatization will do, but perhaps there could be some other finite axiom set which would work. How do we rule that out?

The only proofs of non-finite-axiomatizability I am acquainted with are for PA and ZF, and they employ reflection schemas. What is the alternative here?

Nagase
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  • Usually, Peano's Axioms are stated in language of set theory as:
    1. $0\in N$
    2. $\forall x\in N: S(x)\in N$
    3. $\forall x,y\in N: [S(x)=S(y) \implies x=y]$
    4. $\forall x\in N: S(x)\neq 0$
    5. $\forall P\subset N: [[0\in P \land \forall x\in P: S(x)\in P]\implies P =N]~~~~$

    Why is this a problem?

     – Dan Christensen Mar 09 '22 at 01:37
  • @DanChristensen - Sorry if I was not clear, but by $\text{Th}(\mathbb{N}, 0, S)$, I meant the first-order theory of the natural numbers with the successor function, not the second-order one. Similarly for the other theories mentioned. – Nagase Mar 09 '22 at 01:54
  • So, if you want finite axiomability (?), you need second-order PA. OK. Thanks. – Dan Christensen Mar 09 '22 at 05:55

2 Answers2

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This is a great example of the power of compactness!

Since first-order logic is compact, it's enough to show that no finite sub-axiomatization of the standard axiomatization given in your earlier question will do the job (see e.g. here). And this is something we can do concretely: letting $\mathcal{M}_n$ be the successor structure consisting of a copy of $\mathbb{N}$ as usual and then a separate $n$-cycle, you can quickly check that each $\mathcal{M}_n$ satisfies the first $n$-many axioms of the theory, but no $\mathcal{M}_n$ satisfies the whole theory.

Note meanwhile that nothing like this can be super useful for $\mathsf{PA}$ or $\mathsf{ZFC}$, each of which have finitely axiomatized subtheories with no "simply-describable" models. And of course the whole idea would break down if we were working in a non-compact logic - showing that a given second-order theory, for example, is not finitely axiomatizable is usually extremely hard since we can't simplify things by focusing on a single well-understood family of candidate finite axiomatizations.

Noah Schweber
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  • Ahh, very nice, I was missing that compactness argument. Many thanks for that, super simple argument. – Nagase Mar 08 '22 at 20:46
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    @tomasz I don't think I agree with that. Take for example second-order logic: we can axiomatize the second-order theory of the ring of integers via a single second-order sentence, or via a fully-non-redundant infinite set of second-order sentences (roughly: "Everything is finitely far from $0$" + "There are at least $n$ elements" for each specific $n\in\mathbb{N}$). In fact, "Every infinite $\mathcal{L}$-theory which is semantically equivalent to a finite $\mathcal{L}$-theory is semantically equivalent to one of its own finite subtheories" is equivalent to "$\mathcal{L}$ is compact." – Noah Schweber Mar 09 '22 at 15:26
  • @NoahSchweber: You're absolutely right. I was really tired, shouldn't have written anything in that state. This is more about covers than about bases, so totally about compactness. – tomasz Mar 09 '22 at 23:49
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Compactness was mentioned in another answer, but compactness is a bit of a red herring here. We can do everything using the $\vdash$ notion of probability and the soundness theorem.

The first step is proving that the axioms of the form $\forall x (S^n(x) \neq x)$ and the axioms $\forall x (0 \neq S(x))$ and $\forall x \forall y (S(x) = S(y) \to x = y)$ are in $Th(\mathbb{N}, 0, S)$. This is pretty straightforward.

The hard part is showing that the above axioms generate $Th(\mathbb{N}, 0, S)$. In other words, we must show that every true statement about $\mathbb{N}$ is provable from the above axioms. I will not go through that here unless this is requested, but this is the critical step.

Next, we show that no finite subset of the above axioms is sufficient to generate $Th(\mathbb{N})$. This involves considering models of the form $\mathbb{N} \cup C_{n + 1}$, where $C_{n + 1}$ is a cycle of length $n + 1$.

Finally, suppose that there is some finite list $a_1, \ldots, a_n \in Th(\mathbb{N})$. Then each $a_i$ can be proved from finitely many of the above axioms, and therefore we can take a finite subset $Q$ of the above axioms which entails all of the $a_i$. But as we we’ve shown above, there is some element of $Th(\mathbb{N})$ which does not follow from $Q$ and therefore does not follow from the $a_i$. So $Th(\mathbb{N})$ is not finitely axiomatisable.

Mark Saving
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    Note that this answer assumes the completeness theorem, which isn't any easier to prove than the compactness theorem; I don't think it's fundamentally different from mine. – Noah Schweber Mar 09 '22 at 01:16
  • @NoahSchweber No, the completeness theorem is not used at all. Only the soundness theorem is required. – Mark Saving Mar 09 '22 at 01:59
  • OK, how do you prove that the above axioms generate $Th(\mathbb{N},0,S)$? The simplest approach I can think of is via the completeness theorem, unless I'm missing something. – Noah Schweber Mar 09 '22 at 04:23
  • @NoahSchweber First, introduce a new function symbol $p$ such that $p(s(x)) = x$ and $p(0) = 0$. The idea is that if $\phi(w, x_1, \ldots, x_n)$ is a quantifier-free formula, then we can produce a quantifier-free $\phi’(x_1, \ldots, x_n)$ and prove using our axioms that $\exists w \phi(w, x_1, \ldots, x_n)$ and $\phi’(x_1, \ldots, x_n)$ are logically equivalent. A bit of induction allows us to prove that any proposition is equivalent to a particular quantifier free proposition over the same variables. In particular, a sentence is equivalent to a quantifier-free sentence. – Mark Saving Mar 09 '22 at 06:23
  • @NoahSchweber Essentially, the key from getting from $\phi$ to $\phi’$ is as follows: let $n$ be the number of occurrences of either $p$ or $s$ in $\phi$. Then it suffices to consider only those $w$ such that either $w \leq n + 1$ or there is some $i$ such that $|w - x_i| \leq n + 1$ (obviously formalised in terms of $0, p, s$). On a separate note, how would you use the completeness theorem to prove that our axioms generate $Th(\mathbb{N})$? – Mark Saving Mar 09 '22 at 06:25
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    Fair - I guess that for me, getting around that sort of argument is one of my favorite things about the compactness theorem (re: "simplest" in my previous comment). Re: using completeness, we can show (e.g. via EF games) that the models of the axiom system in question are all elementarily equivalent to $\mathbb{N}$, and then the completeness theorem says exactly that that means our axioms generate $Th(\mathbb{N}$). – Noah Schweber Mar 09 '22 at 06:27