Let $\sigma = \{E\}$ be a signature with a two-digit relation symbol $E$. Determine for the following $FO-[\sigma]$ formula, whether it has a finite model and whether it has an infinite model.
$$ϕ := ∀x∃y(E(x, y) ∧ ∀z(z \neq y → ¬E(x, z))) ∧ ∀x∀y∀z(E(x, z) ∧ E(y, > z) → x = y) ∧ ∃x∀y¬E(y, x)$$
I am quite frankly lost and not sure what to do here. I know that if $ϕ$ has an arbitrarily large number of finite model, then it has an infinite model. So first I need to determine if $ϕ$ has a model - which would imply that $ϕ$ has a finite model. Then I have to determine if $ϕ$ has an arbitrarily large number of finite models - which would imply that $ϕ$ has an infinite model. But that is where I am not sure what to do. How can I show that $ϕ$ has a model and subsequently that it has an arbitrarily large number of models?
First, a correction: Having a model does not imply having a finite model. There are formulas having only infinite models (try describing how $<$ works on $\mathbb{N}$).
Then, while it is true that having finite models of arbitrarily large cardinalities implies having an infinite model, this is very rarely the way to go to prove that a concretely given formula has an infinite model.
The most basic way to prove that a formula has a finite/infinite model is to just exhibit one. So have a look at what your formula says, and try to construct a model. You can assume without limitation of generality that the underlying set is either $\{0,1,\ldots,n\}$ or $\mathbb{N}$. Then you only need to say how $E$ works.
If you get stuck trying to built a finite model, turn the obstacle you keep running into into a proof that any model has to be infinite.
If you get stuck trying to built an infinite model, then it may help finding out the maximal size of a finite model, and to argue that no larger models can exist.