Elementary equivalence between a well-ordering and linear ordering Consider a language $L$ consisting of a binary predicate $R$ and some additional predicates (e.g. $=$) so that the $L$-structure $M=(U_{M},R_{M})$ is isomorphic to $(\mathbb{N},<_{\mathbb{N}})$.
I am trying to construct an $L$-structure $M'=(U_{M'},R_{M'})$ such that

*

*$M\equiv M'$ (that is, $M$ and $M'$ satisfy the same $L$-sentences)

*$M'$ contains an infinite descending sequence, i.e. there exist elements $a_{1},a_{2},\ldots,a_{n},\ldots\in U_{M'}$ such that $\langle a_{n+1},a_{n}\rangle\in R_{M'}$ for all $n\in\mathbb{N}$.

My first thought was to try with $M'=(\mathbb{Z},<_{\mathbb{Z}})$, but this fails since $M$ does not satisfy the $L$-sentence $\forall x \exists y Ryx$.
Then I considered $R_{M'} = R_{M}\cup\{\langle a_{k+1},a_{k}\rangle : k\in\mathbb{N}\}$, where these $a_{k}$ are new constants. If we consider any finite subset of $R_{M'}$, by an appropriate map we can show that it is satisfiable in the structure $M$ (as $M$ as descending sequences of arbitrary finite length). Then by the compactness theorem $R_{M'}$ is satisfiable is some $L$-structure $M'=(U_{M'},R_{M'})$. Here is where I begin to have troubles:

*

*Is $U_{M'}$ simply the union of $U_{M}$ (i.e. $\mathbb{N}$) and the set of new constants $\{a_{k}:k\in\mathbb{N}\}$?

*Is this $M'$ just obtained by the Compactness theorem and $L$-structure? Since I have introduced constants this is not altogether clear to me because of the following question.

*In trying to show that $M\equiv M$ I run into sentences with the new constants. But these are not $L$-sentences. Or do I have to consider a language $L'=L\cup\{a_{k}: k\in\mathbb{N}\}$? That is, how do I prove that $M$ and $M'$ satisfy the same $L$-sentences?

Thank you.
 A: I found the following MSE related questions: This and this appear directly related to Berci's comment to my question. Then there is this, this, and this.
In this and this Alex Kruckman, whose comments I find quite helpful, provides good answers.
The gist of the argument is to find a set $S$ of sentences (in the language $L$ with parameters $V$) which contain the set of all valid $L$-sentences in $M$ (so that any $L$-$V$-structure's reduct to $L$ without the parameters will be elementary equivalent to $M$), but which also includes sentences describing a phenomenon which does not happen in $M$. We then show that any finite subset of $S$ is satisfiable and the Compactness Theorem takes care of the rest.
Consider the set $T$ of all $L$-sentences $A$ such that $M\vDash A$. Also, consider a set of new parameters $V=\{a_{k}:k\in\mathbb{N}\}$, i.e. not occurring in $M$. Then define the set
$$S=T\cup\{Ra_{k+1}a_{k}:k\in\mathbb{N}\}.$$
We will show that $S$ is finitely satisfiable. Indeed, consider a finite subset $S_{0}\subset S$. Then, the set $C$ of parameters mentioned in the expressions contained in $S_{0}$ is finite. let $\varphi:V\rightarrow U_{M}$ be a map sending $C$ to an appropriate finite descending sequence of elements of $U_{M}$ in the sense that $\langle\varphi(a_{k+1}),\varphi(a_{k})\rangle\in R_{M}$ whenever $Ra_{k+1}a_{k}\in S_{0}$ (such a $\varphi$ can always be found since $U_{M}$ has descending sequences of arbitrary finite length). Clearly the $L$-$V$-structure $(M,\varphi)$ satisfies $S_{0}$.
By the Compactness Theorem there is an $L$-$V$-structure $(M',\phi)$ satisfying $S$. Note that $M'=(U_{M'},R_{M'})$ is an $L$-structure. Moreover, for any $L$-sentence $A$,
$$M\vDash A\Rightarrow A\in T\Rightarrow M'\vDash A.$$
On the other hand,
$$M\nvDash A\Rightarrow M\vDash(\neg A)\Rightarrow (\neg A)\in T\Rightarrow M'\vDash(\neg A)\Rightarrow M'\nvDash\neg A.$$
Therefore $M\equiv M'$. In particular, $M'$ is a linear ordering because the $L$-sentences 1-5 below are satisfied by $M'$.

*

*$\forall x\;\neg Rxx$.

*$\forall x\forall y\;Rxy\Rightarrow\neg Ryx$.

*$\forall x\forall y\forall z\;Rxz\wedge Ryz\Rightarrow Rxz$.

*$\forall x\forall y\forall z\;Rxz\Rightarrow Rxy\vee Ryz$.

*$\forall x\forall y\;(\neg Rxy) \wedge (\neg Ryx) \Rightarrow x=y$.

Finally, note that $M'$ contains an infinite descending sequence since $(M',\phi)\vDash S$ means that $M'\vDash R\phi(a_{k+1})\phi(a_{k})$ for all $k\in\mathbb{N}$, and this means that for the elements $\phi(a_{1}),\phi(a_{2}),\ldots,\phi(a_{n}),\ldots\in U_{M'}$ we have that $\langle\phi(a_{k+1})\phi(a_{k})\rangle\in R_{M'}$, as desired.
Note that $M$ is a strict well-ordering while $M'$ is not. Therefore, $M$ and $M'$ are not isomorphic $L$-structures.
