Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.)

share|cite|improve this answer
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 $\mathcal{Z} \models \varphi (m_2,n_2)$. – arjafi Nov 16 '12 at 15:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.