Proving truth in a model for arbitrary sentences

Question

Define a model $\mathcal{M} := (M, \mathcal{I}, w)$ where $M$ is the domain, $\mathcal{I}$ the interpretation functions and $w$ the truth valuation\assignment.

Also define an arbitrary sentence $\psi (x)$ where $x\in A$ for some set A.

Can you prove that there exists a variable assignment $s(x)$ that makes $\psi (x)$ true in $\mathcal{M}$ (as in $\mathcal{M} \models \psi (x)$) without specifically knowing what the variable assignment $s(x)$ is? Must you prove that with the given interpretations $\mathcal{I}$ in $\mathcal{M}$ it must be true that $x\in M$ assuming $\psi(x)$? It feels intuitively true if you look at simple sentences, for example ($\exists n\in \mathbb{N} : n < 0)$ will be false in the model $(\mathbb{N}, +, x, 0)$ since then ($n\notin \mathbb{N}$) due to the construction of $\mathbb{N}$, but for ($\exists z\in \mathbb{Z} : z < 0$) ($z\in \mathbb{Z}$) can be true in the model $(\mathbb{Z}, +, -, x, 0)$ due to the construction of $\mathbb{Z}$. We don’t know the exact variable assignment but if we can prove $(z\in \mathbb{Z})$ is true where $(z < 0)$ and $(z\in \mathbb{Z})$ are assumed with the given interpretations then shouldn’t $(\exists z\in \mathbb{Z} : z < 0)$ be true in the model $(\mathbb{Z}, +, -, x, 0)$? but i’ve never seen a website or book explicitly state that you should eventually prove that $x$ should be an element of the domain $M$ of $\mathcal{M}$, so i assume there is more to it, but i’m not sure how i should prove truth in a model for some sentence $\psi (x)$ then. (If it depends on the “type” of sentence please explain what this “type” of sentence should be.)

examples can help!

Thank you for reading my question! (I know i’m once again a bit vague so if you need clarification for a specific part if my question please ask for it ;) Also of you know very closely related question please link it in the chat).

Answer 1

$x$ is a variable, a syntactic object . If we want to assign to it some value wrt the domain M we have to use variable assignment function $s: \text {Var} \to M$ (and not a truth assignment).

Example with formula: "x is Even". For an interpretation with domain $\mathbb N$ we have that it is true for the assignment $s(x)=2$.

The same for $\exists x (x+4=8)$. With $s(x)=4$ we have that the formula $(cx+4=8)$ is satisfied in $\mathbb N$ by the variable assignment $s$. And this means that the formula $\exists x (x+4=8)$ is true in $\mathbb N$:

$\mathbb N,s \vDash \exists x (x+4=8)$.

Edit: removed comment wrt previous point (2).

If we modify the formula to e.g. $\exists x (x+4=0)$, we know that there is no number in $\mathbb N$ that satisfies the equation, and thus we cannot define a suitable $s$.

We need a different interpretation, with $\mathbb Z$ as domain and in this case we can find $s'(x)=-4$ such that:

$\mathbb Z,s' \vDash \exists x (x+4=0)$.