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.


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*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?

*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 $T_0 \subseteq T_1$?

*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.
 A: 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.
