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I have to prove the following statement:

Prove that there is no formula $\psi=\psi(x_0,x_1)$ in the language $\operatorname{Th}((\mathbb{Z},S))$ such that the relation $\{(m,n)\in\mathbb{Z}\times\mathbb{Z}: (\mathbb{Z},S)\models \psi[m,n]\}$ is a linear ordering of $\mathbb{Z}$. Conclude that the relation $<$ on $\mathbb{Z}$ is not definable in the structure $(\mathbb{Z},S)$

Notice that $S$ is the successor function and $\psi(x_0,x_1)$ a formula with two free variables. Can someone help me with this question because I have no idea how to solve this problem. I thought about quantifier elimination, but can you solve this also without quantifier elimination, for example with automorphisms?

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1 Answer 1

To use automorphisms to show no formula $\psi ( x,y )$ over the language $\{ S \}$ defines a linear order over $\mathcal{Z} = ( \mathbb{Z} , S )$, note, first, that since the only automorphisms of $\mathcal{Z}$ are the shifts we will not be able to work with $\mathcal{Z}$ itself, but rather some elementary extension.

Let's consider the structure $\mathcal{M} = ( \mathbb{Z} \cup \mathbb{Z}^* , s )$ where

  • $\mathbb{Z}^*$ is a "starred" copy of $\mathbb{Z}$ (disjoint from $\mathbb{Z}$);
  • $s : \mathbb{Z} \cup \mathbb{Z}^* \to \mathbb{Z} \cup \mathbb{Z}^*$ is the naturally defined successor operator: $$\begin{align} s ( n ) &= n+1 \\ s ( n^* ) &= (n+1)^*.\end{align}$$

It is fairly straightforward to show that $\mathcal{Z} \prec \mathcal{M}$, and so if $\psi (x,y)$ defines a linear order on $\mathcal{Z}$ it also defines a linear order, $\sqsubset$, on $\mathcal{M}$. Since the mapping $\sigma : \mathbb{Z} \cup \mathbb{Z}^* \to \mathbb{Z} \cup \mathbb{Z}^*$ defined by $$\begin{align} \sigma( n ) &= n^*\\ \sigma( n^* ) &= n\end{align}$$ is an automorphism of $\mathcal{M}$, it follows that for all $a,b \in \mathbb{Z} \cup \mathbb{Z}^*$ we have $$\mathcal{M} \models a \sqsubset b \quad \Leftrightarrow \quad \mathcal{M} \models \sigma(a) \sqsubset \sigma (b).$$ From this it is easy to show that both of the assumptions $0 \sqsubset 0^*$ and $0^* \sqsubset 0$ lead to contradictions.

(You can also see some previous related questions here and here.)

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Thank you ... i will think about this. But is this also possible with QE? –  Countable Universal Nov 16 '12 at 12:47
    
@CountableUniversal: I believe that quantifier elimination should work, provided you can show that $\mathrm{Th} ( \mathbb{Z} , S )$ admits quantifier elimination. I might be incorrect on some specifics (or even some generalities) but if $\varphi (x,y)$ is quantifier-free then there should be a bound $N$ (likely one more than the number of occurrences of $S$ in $\varphi$) such that given $m_1,n_1,m_2,n_2 \in \mathbb{Z}$ if $| m_1 - n_1 | , | m_2 - n_2 | \geq N$, then $\mathcal{Z} \models \varphi (m_1,n_1)$ iff $\mathcal{Z} \models \varphi (m_2,n_2)$. –  Arthur Fischer Nov 16 '12 at 15:18

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