Showing a model does not have a particular substructure and understanding satisfaction relations. Out of Winfried Just and Martin Weese's Set theory book:

Show that the model $\mathfrak B=\langle \Bbb Z, +, \le, 0 \rangle$ does not have any substructure whose universe is $\{-1,0,1\}$.

In a previous example he showed that there is such a substructure for $\mathfrak B$ with multiplication instead of addition. However, I do not see how changing the function breaks the identity embedding. My guess is that it has to do with {-1,0,1} not being closed under addition but I am unsure how to formalize this. Something like $i(1+1)\neq i(1)+i(1)$ since $1+1\notin \{-1,0,1\}$ where $i$ is the identity function is my attempt but this is clearly lacking.
Also on a somewhat related note, I'm not quite sure I understand the following property of satisfaction relations under a valuation s

$\mathfrak U \vDash_s \exists v_i\phi$ iff there exists a valuation $s^*$ such that $\mathfrak U \vDash_s^* \phi$ and $s(k)=s^*(k)$ for all $k\neq i$

How I understand this property is that a bounded variable's formula is true under the valuation $s$ as long as the formula can be satisfied under a valuation for all of the free variables of $\phi$ in $\mathfrak U$. Is this interpretation correct?
 A: You have the right idea: any substructure of a structure must be closed under all functions of the structure. The way to formalise it is to assume there is a structure $\cal A$ with domain $\{-1, 0, 1\}$. But then ${\cal A} \models 1 + 1 = -1$ or ${\cal A} \models 1 + 1 = 0$ or ${\cal A} \models 1 + 1 = 1$. In all cases the same does not hold in $\cal B$. Hence $\cal A$ is not a substructure of $\cal B$.
As for the second question, it is a formal definition, so I can't explain it more formally. However, I'll try to give some intuition behind it. Valuations are assigning elements of the domain to variables. They assign values to all variables (that is more convenient technically), but what matters are the values assigned to the free variables of the formula. Now $\cal A \models_s \exists v_i \phi$ is defined to hold if there is an element $a$ in the domain of $\cal A$ such that if you expand (exchange) your assignment with assigning $a$ to $v_i$ then the formula $\phi$ will be true for that assignment. Note that the variable $v_i$ is not free in $\exists v_i \phi$. So the value of $v_i$ under $s$ is not important.
