Is this logical statement correct when the universe is the set of integers?

Question

∀x∃y((y > x) ∧ ∀z(((z $\ne$ y) ∧ (z > x)) ⇒ (z > y)))

I'm confused by the brackets. Right now I'm leaning towards the statement being true as it says that for all x we can find a y, and z such that the z is greater than y and x.

Answer 1

$\forall x~\exists y~\color{red}{\Big(}\color{blue}(y > x\color{blue}) \land \forall z~\color{purple}{\bigl(}\color{limegreen}(\color{pink}(z ≠ y\color{pink}) \land \color{fuchsia}(z > x\color{fuchsia})\color{limegreen}) \to \color{gold}(z > y\color{gold})\color{purple}{\bigr)}\color{red}{\Bigr)}$

"For any integer $x$ there is some integer $y$ that is greater than $x$, and every integer $z$ will be greater than $y$ if it is both not equal to $y$ and greater than $x$."

So, the claim is that: You can select any integer and will be able to find some second integer that is greater than the first but with no third integer that can be found between them.

Or simply "Every integer has a least greater integer."