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Let $\mathcal{L}=\{\leq\}$ and $\mathcal{A}=\{\mathbb{N},\leq\}$, and $\mathcal{B}=\{K,\leq\}$ where $K=\{(a,b)\in\mathbb{N}^2:a\in\mathbb{N} \text{ and } b=1 \text{ or } b=2\}$ and pairs $(a,b)$ have the natural ordering on themselves but pairs $(a,2)$ are strictly greater then all pairs of the form $(a,1)$

Show that $\mathcal{A}\not\equiv\mathcal{B}$

So I believe what I want is a sentence which is true in one structure but not true in the other structure.

I can obviously identify some things that are properties of the naturals that won't be properties of the set $K$ like every natural has only finitely many elements less then or equal to it. But not ones that can be expressed in a single sentence.

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    $\begingroup$ In $\mathbb N$ we have only one number that has no predecessor while in $K$ we have two : $(0,1)$ and $(0,2)$. $\endgroup$ – Mauro ALLEGRANZA Oct 16 '19 at 15:08
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    $\begingroup$ Right, but without constant $1$, and the $+$ operation in $\mathcal{L}$ can this be expressed as a sentence? $\endgroup$ – AColoredReptile Oct 16 '19 at 15:09
  • $\begingroup$ What does it mean to be a predecessor? $x$ is a predecessor of $y$ if $x$ is the biggest thing less than $y$. Can you express this with only $<$? $\endgroup$ – HallaSurvivor Oct 16 '19 at 15:11
  • $\begingroup$ Ok $\neg\exists_z(x\leq z \wedge z\leq y)$ $\endgroup$ – AColoredReptile Oct 16 '19 at 15:17
  • $\begingroup$ @MauroALLEGRANZA That doesn't work either: your $z$ could be $m$ or any element less than $n$. You want $n\leq m \land n\neq m \land \forall z( (n\leq z \land z\leq m)\rightarrow (z = n\lor z = m))$ $\endgroup$ – Alex Kruckman Oct 16 '19 at 17:05
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Exactly one element of $\mathcal{A}$ has no predecessors, while two distinct elements of $\mathcal{B}$ have no predecessors. See if you can write $\text{isPred}(x,y)$ and use this to distinguish the structures.


I hope this helps ^_^

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    $\begingroup$ Thanks I have a good idea of what to do now. $\endgroup$ – AColoredReptile Oct 16 '19 at 15:19

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