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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?

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

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