Logic: defining 2 in a certain structure

Question

Let $L= \{ \leq \}$ be the language of the partial orders and $M$ the $L$-structure with $M=\{ 1,2,3,4,6,12\}$ and $\leq_M=\{(x,y)$: $x$ is a divisor of $y$$\}$. Now, give an $L$-formula $\phi(x)$ that defines the element $2$ (or equivalently, for every $m \in M$: $M \models \phi [m/x] \Leftrightarrow m=2$).

Is this a good formula $\phi$? $\quad$ $ \neg(x \leq_M 1) \wedge (x \leq_M 2) \wedge \neg (x \leq_M 3) \wedge (x \leq_M 4) \wedge (x \leq_M 6) \wedge (x \leq_M 12) $

Answer 1

No, the proposed formula is not a good formula. The reason is that your language only includes $\leq$, and you are using $1,2,3,4,6$ and $12$. If you're going at it this way, you can just take $x=2$ and get it over with.

The correct way, instead is to pinpoint a property expressible by $\leq$, as interpreted in $M$. For example, you can note that $2$ is "almost" the minimum, and that except the minimum, and another element, every other number in $M$ is divisible by $2$. Try to formalize this, and prove that indeed only $2$ satisfies both these properties.

Answer 2

Hint:

• The brutal solution is like this: \begin{align} \phi(x) &= \exists m_1,m_2,m_3,m_4,m_6,m_{12}.\ x = m_2 \\ &\land \left(\bigwedge_{i \neq j} m_i \neq m_j\right) \land \left(\bigwedge_{i \text{ is a divisior of } j} m_i \leq_M m_j\right) \land \left(\bigwedge_{i \text{ is not a divisor of } j} m_i \not\leq_M m_j \right) \end{align}
• Observe that I do not use elements of $M$ in it (the indices of $m$ are different things and only "coincide" with elements of $M$ so that it would be easier for me to write $\phi(x)$, but it could have start with $\exists m_\spadesuit, m_\heartsuit,m_\diamondsuit,m_\clubsuit,m_\bigstar,m_\maltese$)
• Try optimizing it for your case.

I hope this helps $\ddot\smile$

Answer 3

A less brutal solution would be to notice that $2$ is the only element in the set that has exactly three strictly larger elements.