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

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?

• Over $\Bbb Z$ can you describe, in terms of $x$, a suitable $y$? – Lord Shark the Unknown Jun 27 '18 at 6:24
• I think it's $x-5<y<x$ – user123429842 Jun 27 '18 at 6:29
• The concept of predecessor may be useful. – AnyAD Jun 27 '18 at 6:35

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 $x - 5 < y < x$).