Is $\exists x \forall y \exists z P(x,y,z)$ satisfiable?

Question

I have this formula: $$\begin{align} \exists x \forall y \exists z P(x,y,z) \\ \end{align}$$

How to check whether it is satisfiable? I know that I have to find a structure in which it is true.

Answer 1

You can find a very "slim" model for the formula :

$∃x∀y∃zP(x,y,z)$,

showing thus that it is satisfiable.

Consider a domain $D = \{ a \}$, i.e. with just one element, and assume that the predicate letter $P$ stands for a relation $P^D$ such that $P^D(a,a,a)$ holds.

Then, clearly, $∃x∀y∃zP(x,y,z)$ is true in $D$.

We can "mimick" this "slim" model with the natural numbers, considering $D = \{ 0 \}$ and again interpreting $P(x,y,z)$ as $x+y=z$.

We have that $0+0=0$, and thus, due to the fact that $0$ is the only element of the domain, $∃x∀y∃z(x+y=z)$ is true in $D$.