Is $\forall x\exists y((y<x)\land (x<y+5))$ true if $x,y \in \mathbb N$ and $x,y\in \mathbb Z$?

Question

Is $\forall x\exists y((y<x)\land (x<y+5))$ true if $x,y \in \mathbb N$ and $x,y\in \mathbb Z$? $\mathbb N$ includes $0$.

I think that if $x,y \in \mathbb N$ then it's not true if $x=0$.

But for $x,y \in \mathbb Z$ I think it is true. Is it correct?

Answer 1

Over ${\mathbb N}$ this is indeed false. In fact, already $\forall x \in {\mathbb N} \; \exists y \in {\mathbb N}:y<x$ is false; $x = 0$ is a counterexample.

Over ${\mathbb Z}$ this is indeed true. Given $x \in {\mathbb Z}$, take $y = x - 1$ (or $x - 2$, $x - 3$, $x - 4$, because, as you noticed, the condition is equivalent to $x - 5 < y < x$).