Expanding a model $M$ into a model $N$ vs. elementarily equating $M$ with a reduct of $N$ Do the following statements mean the same thing? If not, what's the difference? Assume that $T_0, T_1$ are first-order languages and that $T_1$ is a model-theoretic extension of $T_0$.

Every model for $T_0$ can be expanded into a model for $T_1$.

vs.

For any model $M$ of $T_0$ there is a model $N$ of $T_1$ such that $M \equiv N\upharpoonright L_0$.

Notes:


*

*"$N\upharpoonright L_0$" means the reduct of $N$ to the language of $M$.

*"$\equiv$" means elementary equivalence, not isomorphism, so that, for instance, $(\mathbb{N},<)\equiv(\mathbb{N}+\mathbb{Z},<)$, as can be proved e.g. via quantifier elimination, but, clearly, $(\mathbb{N},<)\not\cong(\mathbb{N}+\mathbb{Z},<)$.

My confusion may arise from an inadequate understanding of the meanings of one or more of the terms model-theoretic extension, expansion, reduct, elementary equivalence and isomorphism.
Please assume in your answer minimal knowledge of logic on my part.
 A: This was explained in my answer to part 3b of your previous question: they are not equivalent (although the first trivially implies the second). Isomorphism (= existence of a structure-preserving bijection) is strictly stronger than elementary equivalence (= satisfy the same first-order sentences), and this drives the separation.
In more detail:
Let $L_0$ be the language of arithmetic, and $L_1$ be that language expanded with a new constant symbol $c$. Let $T_0$ be the theory of $(\mathbb{N}; +, \times, 0, 1)$, and $T_1$ be that theory together with the sentences $$\mbox{"$c>1+1+1+...+1$"}$$ ($n$ "1"s) for each $n\in\mathbb{N}$.  Then:


*

*Clearly $T_1$ is a model-theoretic (in fact, proof-theoretic) extension of $T_0$.

*For any model $N$ of $T_1$ and any model $M$ of $T_0$, $M\equiv N$. (This is because $T_0$ is a complete theory in the language $L_0$.)

*But the model $(\mathbb{N}; +, \times, 0, 1)$ has no expansion to a model of $T_1$: wherever we "put" $c$, we wind up violating one of the axioms of $T_1$.
(Note, to allay possible worries, that $T_1$ does indeed have models - this can be proved using the compactness theorem, and is a good exercise.)

You asked in a comment to bof's answer how to tell that two structures are elementarily equivalent. There are several ways to do this:


*

*If (the deductive closure of) $S$ is a complete $L$-theory and $A, B\models S$ are $L$-structures, then $A\equiv B$ by definition: if $A\models \varphi$, then we can't have $\neg\varphi\in S$, so since $S$ is complete we have $\varphi\in S$, but then $B\models\varphi$ since $B\models S$.

*We can use Ehrenfeucht-Fraisse games to show that two structures are elementarily equivalent.

*Los' Theorem says that we can show that two structures are elementarily equivalent if they have isomorphic ultrapowers (the converse is the Keisler-Shelah isomorphism theorem).

*Quantifier elimination, model completeness, or similar can be used to give a complete description of the theory of a structure, and so facilitate proofs of elementary equivalence.

*And sometimes you can come up with really silly model chain constructions too.
And there are lots of other methods.
