# Proof- vs. model-theoretic definitions of extension and of conservative extension

Let $L_0, L_1$ be first-order languages, such that $L_0$'s signature is a subset of $L_1$'s signature, in the sense that, for all $n\in\{0,1,2,\dots\}$: every $n$-ary function/predicate of $L_0$ is also an $n$-ary function/predicate of $L_1$, respectively (constants are regarded as $0$-ary functions). The sentences of a first-order language are its well-formed formulas with no free variables. Let $T_0$ be a set of $L_0$-sentences, and let $T_1$ be a set of $L_1$-sentences.

1. Consider the following proof-theoretic definition of extension.

$T_1$ is an extension of $T_0$ iff $T_0\subseteq T_1$.

Now consider the following model-theoretic definition of extension.

$T_1$ is an extension of $T_0$ iff $T_1 \vDash T_0$ (i.e., if every model of $T_1$ is also a model of $T_0$), when $T_0$ is regarded as an $L_1$-theory.

Are these definitions equivalent? I can see that the proof-theoretic definition implies the model-theoretic one, but does the converse hold? If these definitions are not equivalent, is it possible to formulate a model-theoretic equivalent to the proof-theoretic definition?

2. Consider the following, proof-theoretic, definition of conservativity.

$T_1$ is conservative w.r.t. $T_0$ iff whenever $\phi$ is an $L_0$-sentence, then $\vdash_{T_1} \phi$ implies $\vdash_{T_0} \phi$.

Is there an equivalent model theoretic definition, in the sense that the definition does not rely on the concept of proof or any concept derived from it (such as provability and the relation $\vdash$)? If not, what if $T_0 \subseteq T_1$?

3. a) Suppose every model of $T_0$ can be expanded into a model of $T_1$. Can it be deduced that $T_1$ is conservative w.r.t. $T_0$? If not, is there some other sufficient model-theoretic condition that guarantees that $T_1$ be conservative w.r.t. $T_0$?

b) Suppose $T_1$ is conservative w.r.t. $T_0$. Can it be deduced that every model of $T_0$ can be expanded into a model of $T_1$? If not, what if additionally $T_0\subseteq T_1$?

Please assume minimal knowledge of logic on my part. Feel free to refer me to textbooks or other sources, but if you do so, try to be as specific as possible. Thanks.

• For Q.1., Consider the case $L_0=L_1$ and $K_1$ is a proper subset of $K_0$ but $K_0$ \ $K_1$ consists of theorems of $K_1.$ – DanielWainfleet Feb 5 '17 at 8:00
• Re: your most recent edit, first of all I don't see how my answer to (2) "relies on the concept of proof or any concept derived from it"; my characterization just used models and satisfaction. Second, "$T\vdash \varphi$" is equivalent to "$T\models \varphi$" (the latter of which means "every model of $T$ is a model of $\varphi$"). The hard direction - "$\models$ implies $\vdash$" - is the Completeness theorem. Do you consider this appropriately model-theoretic? – Noah Schweber Feb 8 '17 at 22:17
• @NoahSchweber: Yes, I do consider it model-theoretic, as it can be defined independently of the concept of proof. The fact that the two can be shown to be equivalent is beside the point. – Evan Aad Feb 8 '17 at 22:19
• @EvanAad OK, so doesn't that answer (2) immediately? Just say "$T_1$ is model-theoretically conservative w.r.t. $T_0$ iff $T_1\models\varphi$ implies $T_0\models\varphi$ for all $\varphi$ in the language $L_0$". (I thought you wouldn't like this, which is why I went to the lengths I did in my answer - the Keisler-Shelah iso. thm. gives a purely model-theoretic characterization of "$\equiv$", which is all I needed. So if I understand your previous comment correctly, I can greatly simplify my answer to (2).) – Noah Schweber Feb 8 '17 at 22:20
• I guess you're right, but part of the difficulty when I posed the problem was that I wasn't as aware as I am now of the equivalence between $\vdash$ and $\vDash$. Please don't simplify your answer. It is very illuminating. How would you define "model-theoretic"? Not the same as I did? – Evan Aad Feb 8 '17 at 22:25

I suspect you meant to include the requirement that $T_1$ be stronger than $T_0$ - that is, a model-theoretic extension - in your definition of conservativity, so I've done so below.

Short version: any definition that makes specific reference to the elements of a theory as opposed to the elements of its deductive closure has no model-theoretic equivalent. Why? Well, a theory $T$ is model-theoretically equivalent to its deductive closure $cl(T)$ (by the Soundness Theorem, they have exactly the same models). So you won't find a model-theoretic property that holds of $T$ but fails of its deductive closure - in particular, statements like "$T_1\supseteq T_0$" are inherently non-model-theoretic, since there are examples of theories $S$ and $T_0$ such that $S\not\supseteq T_0$ but $T_1=cl(S)\supseteq T_0$.

In particular, anytime $cl(S)=T_1\not=S$, we have a counterexample to (1): $S$ is model-theoretically an extension of $T_1$ (every model of $T_1$ is a model of $S$), but not proof-theoretically.

We now have two ways to define "conservative over". The first ultimately turns out to be model-theoretic:

Definition 1. $T_1$ is a model-theoretic conservative extension of $T_0$ if for every $\varphi$ in the language of $T_0$, $T_1\vdash\varphi$ iff $T_0\vdash\varphi$.

This has an equivalent formulation:

Fact 1. Suppose $T_1$ is a model-theoretic extension of $T_0$. Then $T_1$ is a model-theoretic conservative extension of $T_0$ iff for any model $M$ of $T_0$, there is a model $N$ of $T_1$ such that $M\equiv N\upharpoonright L_0$ (where "$N\upharpoonright L_0$" is the reduct of $N$ to the language of $M$); and conversely.

Proof: Suppose the property in Fact 1 holds. If $T_1$ weren't a model-theoretic conservative extension of $T_0$, then there would be some sentence $\varphi$ in the language $L_0$ such that $T_1\vdash \varphi$ but $T_0\not\vdash\varphi$. Let $M\models T_0\cup\{\neg\varphi\}$; then $M$ can't be elementarily equivalent to the reduct of any model of $T_1$, contradicting the Fact 1 property.

Conversely, if Definition 1 holds, then let $M_0\models T_0$ and consider the theory $T_1\cup Th(M_0)$. This theory is finitely consistent: otherwise, there would be some $\psi\in Th(M_0)$ such that $T_1\vdash\neg\psi$, but then - since $T_0\not\vdash\neg\psi$ - this would contradict Def. 1. But then by Compactness, $T_1\cup Th(M_0)$ has a model $N$; and then it's obvious that $N\upharpoonright L_0\equiv M_0$ (they have the same $L_0$-theory by definition); so the property in Fact 1 holds. Similarly, if $T_1$ proves every $L_0$-sentence that $T_0$ proves, then the reduct of any model of $T_1$ is a model of $T_0$. $\Box$

Fact 1 gives a characterization of conservativity in terms of elementary equivalence. But here's the thing: is elementary equivalence a model-theoretic proprety? On the one hand, it's definitely a property of structures; on the other hand, it makes reference to sentences. A very restrictive definition of "model-theoretic" might seem to exclude it.

However, it turns out that we can get around this! The point is that there is a completely model-theoretic characterization of elementary equivalence (this is the Keisler-Shelah isomorphism theorem). With this in hand, we have an immediate model-theoretic characterization of model-theoretic conservativity:

Fact 2. $T_1$ is a model-theoretic conservative extension of $T_0$ iff $T_1$ is a model-theoretic extension of $T_0$ (the reduct of any model of $T_1$ is a model of $T_0$) and for any model $M$ of $T_0$, there is some $N\models T_1$ such that $M$ and the reduct of $N$ have isomorphic ultrapowers.

Note that it doesn't matter exactly what an "ultrapower" is, just that it is a construction on the model-theoretic side. This is a definition entirely in terms of structures (sentences are never mentioned), so it's purely model-theoretic.

The previous section answered the first part of (2) in the affirmative. However, if we add to conservativity the requirement that $T_1\supseteq T_0$, then the answer is no, there is no model-theoretic characterization. This is just because of what I wrote in my first paragraph: any definition that makes explicit reference to the theory rather than its deductive closure has no not model-theoretic equivalent. And the requirement "$T_1\supseteq T_0$" is of this form.

Again, we can use the example of two versions of the same theory to show this. Take $cl(T_0)=T_1$ with $T_0\not=T_1$. Then $T_1$ is a model-theoretic conservative extension of $T_0$, but not proof-theoretically; and no model-theoretic property will distinguish between $T_1$ and $T_0$.

Based on your comments, I think you're unsatisfied with this, but I genuinely don't know why; can you explain what I can do better here?

Finally, we come to (3).

The answer to (3a) is yes, and this is essentially contained in the proof of the Fact above. Namely, suppose every model of $T_0$ can be expanded to a model of $T_1$. Then if $T_1\vdash \varphi$, $\varphi$ is true in every model of $T_1$; so it must be true in every model of $T_0$, since if $M\models T_0\cup\{\neg\varphi\}$ then $M$ clearly can't be expanded to a structure in which $\varphi$ is true.

The answer to (3b), meanwhile, is no, and the easiest proof I know makes use of Compactness - specifically, of non-standard models of arithmetic. For example, consider the theory $TA$ of true arithmetic (that is, $Th(\mathbb{N}; +, \times, 0, 1, <)$). This is our $T_0$. Now for our $T_1$, take $T_0$ and add a new constant symbol $c$ to the language, and axioms stating $$c>1, c>1+1, c>1+1+1, . . .$$ Every finite subset of $T_1$ is satisfiable (just interpret $c$ as a big enough natural number), so by Compactness $T_1$ has a model; but any model of $T_1$ has to have an infinite element! So $(\mathbb{N}; +, \times, 0, 1, <)$ is a model of $T_0$ which can't be expanded to a model of $T_1$. Yet it's clear that $T_1$ is conservative over $T_0$.

I hope this answers your question; please let me know if there's anything I can explain further.

• In the penultimate paragraph: could you have meant "there is some $M\vDash T_0\cup\{\neg\varphi\}$" rather than "there is some $M\vDash T_0\cup\{\varphi\}$"? – Evan Aad Feb 5 '17 at 7:32
• By the way, I have reworded my question more crisply. The new rewording does not, I think, affect your answer in any meaningful way (at least not as far as your answers to questions 1 and 3 are concerned), but some slight changes may need to be done. – Evan Aad Feb 5 '17 at 8:37
• I have added another question, 3b. Would you mind taking a look? – Evan Aad Feb 5 '17 at 9:42
• @EvanAad See my edit; the answer is no. – Noah Schweber Feb 5 '17 at 14:17
• Thank you. So this settles questions 1 and 3. There remains question 2. You claim in your post that the answer to 2 is no "for exactly the same reasons as" the ones used to answer question 1, but I don't see how. – Evan Aad Feb 5 '17 at 15:47

Noah gave an excellent answer. His answer uses many standard concepts and results from the theory of logic that I had to look up. In order to make future revisits easier, I will rewrite his answer with greater detail and with references to the literature where all the relevant definitions and propositions can be found. All references pertain to Peter G. Hinman's Fundamentals of Mathematical Logic (A K Peters 2005), which will be abbreviated as FML.

1. No, the definitions are not equivalent.

Choose $L$ to be any first-order language and choose $\Gamma$ to be any set of $L$-sentences, such that $\Gamma\subsetneq\text{Th}(\Gamma)$, where $\text{Th}(\Gamma)$ [FML Definition 2.4.4, p. 140] is the set of all $L$-sentences $\sigma$ such that $\Gamma\vDash\sigma$, i.e. such that $\sigma$ is true in every model of $\Gamma$. (To see that it is possible to choose such $L$ and $\Gamma$, recall that, by the Completeness theorem [FML 3.4.1(a), p. 237], $\Gamma\vDash\sigma$ iff $\Gamma\vdash\sigma$, i.e. $\Gamma\vDash\sigma$ iff $\sigma$ is provable from $\Gamma$. So we may choose, for example, $L$ to be the language with the signature $\{1,\cdot\}$, and $\Gamma$ to be the set of group axioms. The rich literature on group theory is evidence to the fact that $\Gamma\subsetneq\text{Th}(\Gamma)$.)

Now define \begin{align} L_0 &:= L \\ L_1 &:= L \\ T_0 &:= \text{Th}(\Gamma) \\ T_1 &:= \Gamma \end{align}

Then, trivially, $T_1\vDash T_0$, but $T_0\not\subseteq T_1$.

2. Yes, there is an equivalent model-theoretic definition: just substitute $\vDash$ for $\vdash$, as follows.

$T_1$ is conservative w.r.t. $T_0$ iff whenever $\phi$ is an $L_0$-sentence, then $\vDash_{T_1} \phi$ implies $\vDash_{T_0} \phi$.

To see that this formulation is equivalent to the proof-theoretic one, recall that by the Completeness theorem [FML 3.4.1(a), p. 237] for every first-order language $L$, every set of $L$-sentences $\Gamma$ and every $L$-sentence $\sigma$ we have $\Gamma\vdash\sigma\iff\Gamma\vDash\sigma$.

(This answer is based on an observation made by Noah in a comment to my original post rather than on his answer.)

3. a) Yes, the deduction is warranted.

Let $\phi$ be a $L_0$-sentence such that $\vdash_{T_1}\phi$. To show that $\vdash_{T_0}\phi$ it suffices, by the Completeness theorem [FML 3.4.1(a), p. 237], to show that $\vDash_{T_0}\phi$, which is to say [FML Definition 2.2.1(iv), p. 96] that $\phi$ is true in every $L_0$-model of $T_0$.

Let $M$ be a $L_0$-model of $T_0$. By assumption $M = N\upharpoonright L_0$ for some $L_1$-model $N$ of $T_1$, where $N\upharpoonright L_0$ is the $L_0$-reduct of $N$ [FML Definition 2.6.3, p. 174], i.e. $N$ is an $L_1$-structure that assigns to every non-logical symbol of $L_0$ the same interpretation as assigned by $M$. Since, by assumption, $\vdash_{T_1}\phi$, the Completeness theorem implies that $\vDash_{T_1}\phi$, which implies [FML Definition 2.2.1(iv), p. 96] that $\phi$ is true in $N$. Hence $\phi$ is true in $M$.

b) No, the deduction is unwarranted, even if we allow for $T_0$ to be a subset of $T_0$.

Suppose, to the contrary, that whenever $T_0\subseteq T_1$ and $T_1$ is conservative w.r.t. $T_0$, then every model of $T_0$ can be expanded into a model of $T_1$. We shall derive a contradiction.

Denote by $L_0$ the first-order language whose non-logical symbols consist of the constant symbol $\overset{\cdot}{0}$ and of the unary function symbol $\overset{\cdot}{S}$. (This is the language $L_{\textbf{Sc},0}$ described in the first paragraph on p. 161 of FML. Note that we follow FML's convention of considering the equality symbol $\overset{\cdot}{=}$ to be a logical symbol of every first-order language [FML Definition 2.1.2(vi), p. 86].)

Define the following $L_0$-sentences: \begin{align} \sigma_1 &:= \forall x(\neg\overset{\cdot}{0}\overset{\cdot}{=}\overset{\cdot}{S}x) \\ \sigma_2 &:= \forall x\forall y(\overset{\cdot}{S}x\overset{\cdot}{=}\overset{\cdot}{S}y\rightarrow x\overset{\cdot}{=}y) \\ \sigma_3 &:= \forall x\big(\neg x\overset{\cdot}{=}\overset{\cdot}{0}\rightarrow\exists y(x\overset{\cdot}{=}\overset{\cdot}{S}y)\big) \\ \end{align} Letting $t$ denote any $L_0$-term, Define $t+0 := t$, and for every $n\in\{1, \dots\}$ define $t+n$ to be the $L_0$-term $\overset{\cdot}{S}\Big(\overset{\cdot}{S}\big(\cdots\overset{\cdot}{S}(t)\big)\Big)$ with exactly $n$ occurrences of the symbol $\overset{\cdot}{S}$. Define the $L_0$-sentence $\phi_n$ as follows: $$\phi_n := \forall x(\neg x\overset{\cdot}{=}x+n)$$ Define $T_0 := \{\sigma_1, \sigma_2, \sigma_3\} \cup \{\phi_n\ :\!|\ n\in\mathbb{N}\}$.

Denote by $M$ the $L_0$-structure whose universe is $\mathbb{N}$ (= the set of non-negative whole numbers), and that assigns $\overset{\cdot}{0}\hspace{0cm}^M:=0$ and $\overset{\cdot}{S}\hspace{0cm}^M := \textbf{Sc}$, where $\textbf{Sc}$ is the successor function $$\textbf{Sc}:\mathbb{N}\rightarrow\mathbb{N},\hspace{1cm}\textbf{Sc}(n) = n + 1$$ (We follow's FML's convention of always interpreting $\overset{\cdot}{=}$ as the true equality relation $=$ on the domain of the structure under consideration [FML Definition 2.1.19(i), p. 91].)

We denote by $\text{Th}(M)$ the set of all $L_0$-sentences that are true in $M$ [FML Definition 2.4.8, p. 141]. Then it can be shown [FML Theorem 2.5.8, p. 164] that $\text{Th}(T_0) = \text{Th}(M)$, i.e. that for every $L_0$-sentence $\sigma$, $T_0\vDash\sigma\iff M\vDash\sigma$. Intuitively, $T_0$ is an axiomatization of the familiar natural number system. Another consequence of the fact that $\text{Th}(T_0) = \text{Th}(M)$ is that $M$ is a model of $T_0$.

Denote by $L_1$ the first-order language that expands $L_0$ by adding the constant $c$ to the non-logical symbols of $L_0$, and for every $n \in \mathbb{N}$ define \begin{align} \Gamma_n &:= \{\neg c \overset{\cdot}{=} \overset{\cdot}{0} + k\ :\!|\ k \in \{0, 1, \dots, n\}\} \\ T_1^{(n)} &:= T_0 \cup \Gamma_n \end{align} We have $\Gamma_0\subseteq\Gamma_1\subseteq\cdots$, and therefore also $T_1^{(0)}\subseteq T_1^{(1)}\subseteq\cdots$. Define \begin{align} \Gamma &:= \{\neg c\overset{\cdot}{=} \overset{\cdot}{0}+k :\!|\ k\in\mathbb{N}\} \\ T_1 &:= T_0 \cup \Gamma \end{align} Then $\Gamma = \bigcup_{n=0}^\infty\Gamma_n$ and $T_1 = \bigcup_{n=0}^\infty T_1^{(n)}$, and so if $S$ is any finite subset of $T_1$, $S\subseteq T_1^{(n)}$ for some $n\in\mathbb{N}$.

If $N$ is any $L_1$-structure that is an expansion of $M$ [FML Definition 2.6.3, p. 174], and if $\alpha$ is any function from the set of $L_1$-variables to $\mathbb{N}$ (which is $M$'s, hence $N$'s, domain), and if $V_{N,\alpha}$ is the valuation function [FML Definition 2.1.17, p. 91], then for every two $L_1$-terms $s,t$ and for every $n \in \mathbb{N}$ we have $$V_{N,\alpha}(s = t + n) = \begin{cases} \textbf{T}, &V_{N,\alpha}(s) = V_{N,\alpha}(t) + n \\ \textbf{F}, &\text{otherwise} \end{cases}$$

Therefore $M$ cannot be expanded into a model $N$ of $T_1$, because if this were the case, given any variable assignment $\alpha$ we would obtain $$V_{N,\alpha}(\phi_{\hspace{0cm}_{V_{N,\alpha}(c)}}) = V_{N,\alpha}\big(\neg c=\overset{\cdot}{0}+V_{N,\alpha}(c)\big) = \big(V_{N,\alpha}(c)\neq 0 + V_{N,\alpha}(c)\big) = \textbf{F}$$ in contradiction to the fact that $\phi_{\hspace{0cm}_{V_{N,\alpha}(c)}} \in T_1$ (by definition of $T_1$).

To finish the proof, we need to show that $T_1$ is conservative w.r.t. $T_0$, because then, by the original assumption, there would be an expansion of $M$ into an $L_1$-model of $T_1$, contradicting the results of the previous paragraph. According to the answer to question 2, we are required to show that whenever $\phi$ is a $L_0$-sentence such that $T_1\vDash\phi$, we have $T_0\vDash\phi$.

Let $\phi$ be a $L_0$-sentence such that $\vDash_{T_1}\phi$. Showing that $\vDash_{T_0}\phi$ is tantamount to showing that for every $L_1$-model $K$ of $T_0$, $\vDash_K\phi$. Let $K$ be any model of $T_0$. By the Compactness theorem [FML 3.3.21(i), p. 228] there is some finite $S \subseteq T_1$ such that $S\vDash\phi$. Choose some $n\in\mathbb{N}$ such that $S\subseteq T_1^{(n)}$. Denote by $N$ the $L_1$-structure that expands $K$ by assigning $c^N := n + 1$. Then $N$ is a model of $S$. Therefore $\vDash_N\phi$. Therefore $\vDash_K\phi$.

• +1 - I've only glanced at this a little, but it looks good, and it's awesome that you took the time to write it up! (This is exactly how to learn complicated math in my opinion.) – Noah Schweber Feb 11 '17 at 3:46