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Let $L$ be a language consisting of a binary relation symbol $R$. Consider the following $L$-structures on $\mathbb N$:

$\mathcal N_1=(\mathbb N, \equiv_2),\quad \mathcal N_2=(\mathbb N,<),\quad\mathcal N_3=(\mathbb N,|),\quad\mathcal N_4=(\mathbb N,E)$
where $m\equiv_2 n$ iff $2|m-n$, $<$ is the usual ordering of natural numbers, $|$ is the usual divisibility relation, and $mEn$ iff $m$ and $n$ are consecutive and the smaller one is even. Show that any two of these structures are not elementarily equivalent to each other.

My attempt:
Definition: $\mathcal M \equiv \mathcal N$ if for every $L$-sentence $\sigma$, we have $\mathcal M\models \sigma \iff \mathcal N\models \sigma$

Take $\mathcal N_1 $ and $\mathcal N_2$:

Let $\varphi:=\exists x \exists y(x=3 \, \wedge \, y=4)$

Now, $\mathcal N_1 \nvDash \varphi$ since $2\nmid 3-4$ but $\mathcal N_2 \models \varphi$ since $3<4$.

So, they are not elementarily equivalent to each other.

I don't know how else argue this.

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  • $\begingroup$ You cannot express 3 or 4 in your language, so that is not a valid sentence. You really have to look at the properties of these relations. For example: the relation $E$ (in $\mathcal{N}_4$) is not transitive, but $<$ (in $\mathcal{N}_2$) is transitive. This can be expressed in a logical formula (how?). $\endgroup$ – Mark Kamsma Jun 10 at 21:31
  • $\begingroup$ $\varphi:=\forall x \forall y \forall z(x<y \, \wedge \, y<z \implies x<z)$ How can I generalize this formula(without plugging $<$ in to the formula like I did here) so I can check $\mathcal N_2 \models \varphi$ but $\mathcal N_4 \nvDash \varphi$ @MarkKamsma $\endgroup$ – Leyla Alkan Jun 10 at 21:54
  • $\begingroup$ That is indeed the right idea. As you already said: the way it is written now only makes sense for $\mathcal{N}_2$. In the exercise there is a binary relation symbol $R$ that is interpreted differently in the 4 structures. So for example, $\mathcal{N}_2$ interprets it as $<$. In other words, replacing $<$ by $R$ in your formula gives you a valid formula that just states "$R$ is transitive". This is then true in $\mathcal{N}_2$, but false in $\mathcal{N}_4$. $\endgroup$ – Mark Kamsma Jun 10 at 22:00
  • $\begingroup$ Thanks, I get it now. So, since $\equiv_2$ is an equivalence relation $\mathcal N_1$ differs from all the other structures already. Using transivity $\mathcal N_2$ differs from $\mathcal N_4$ and similarly $\mathcal N_3$ differs from $\mathcal N_4$. Also since $<$ is a linear ordering $\mathcal N_2$ differs from $\mathcal N_3$. $\endgroup$ – Leyla Alkan Jun 10 at 22:22
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    $\begingroup$ That sounds right. Now all you have to do is formalise these ideas using formulas (as you did for transitivity), and then you have your proof! $\endgroup$ – Mark Kamsma Jun 10 at 22:26
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Since your language has no constants you cannot use $3$ or $4$ the way you did there. By finding a sentence that is valid in one but not in other structure, these two structures shall not be elementarily equivalent.

First, observe that $\mathcal N_1$ and $\mathcal N_4$ satisfy the symmetric property, whereas $\mathcal N_2$ and $\mathcal N_3$ do not. Fortunately, this property can be expressed in this language as the sentence $\varphi:= \forall x\forall y\, xRy \to yRx$. So, we only need to show that $\mathcal N_1$ and $\mathcal N_4$ are not elementarily equivalent and that $\mathcal N_2$ and $\mathcal N_3$ aren't either.

For $\mathcal N_1$ and $\mathcal N_4$, observe that $\mathcal N_1\models \exists x\exists y \forall z (zRx\vee zRy)$, i.e. there are two natural numbers that are not related with one another and such that any other number shall be related with one of them. It happens that $\mathcal N_4$ does not satisfy this sentence, by construction of the relation $E$. (this part has been edited following the very useful comment by @Mark Kamsma)

It only remains to be shown that $\mathcal N_2$ and $\mathcal N_3$ are not elementarily equivalent. To this end, observe that $\mathcal N_2$ is a total order while $\mathcal N_3$ is not. The total order formula in this language is $\psi:= \forall x\forall y\, xRy\vee yRx$; it happens that $\mathcal N_2\models \psi$ while $\mathcal N_3\nvDash \psi$.

I hope this helps.

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  • $\begingroup$ Addendum: I have just read your conversation with @Mark Kamsma, which wasn't there while I was writing. That is pretty much what you have to do. $\endgroup$ – Sam Skywalker Jun 10 at 22:36
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    $\begingroup$ Always useful to still have an answer, for people that find this question later on. Although the sentence you used for $\mathcal{N}_1$ versus $\mathcal{N}_4$ is still true in $\mathcal{N}_4$. Given $x$, if $x$ is even (that includes $x=0$) we take $y=x+1$, otherwise we take $y=x-1$. What we can do though, is one of the following "for every $x$ there are at least two different elements related to it" or "there are $x$ and $y$ such that every element is related to $x$ or $y$" (personally I think the second one is the clearest, but they should both work perfectly fine). $\endgroup$ – Mark Kamsma Jun 11 at 7:40
  • $\begingroup$ You are spot on, thanks for pointing it out. I am editing my answer. $\endgroup$ – Sam Skywalker Jun 11 at 7:53
  • $\begingroup$ I could also have opted for the transitive property, but it didn't come to my mind. Really, what was I thinking about to not realise that I was implying that $0$ is not even? $\endgroup$ – Sam Skywalker Jun 11 at 8:03
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    $\begingroup$ We all have those moments ;) In your current answer the bit $x \not R y$ is not really necessary, but it is also not wrong. So I upvoted anyway. $\endgroup$ – Mark Kamsma Jun 11 at 8:07

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