Proving/disproving $\exists n {\in} \mathbb Z \;\forall k {\in} \mathbb Z \;\exists d {\in} \mathbb Z \;\;k+ n = 2d$

Question

I am struggling to wrap my head around what this multiply quantified statement means:

$$\exists n {\in} \mathbb Z \quad \forall k {\in} \mathbb Z \quad \exists d {\in} \mathbb Z \quad k+ n = 2d.$$

Interpreting it as ‘There exists an integer $n$ which for all integers $k$ there exists another integer $d$ where $k+n=2d$’, I think it's true? Where to even begin to (dis)prove this statement?

Answer 1

$\exists n {\in} \mathbb Z \quad \forall k {\in} \mathbb Z \quad \exists d {\in} \mathbb Z \quad k+ n = 2d.$

Interpreting it as ‘There exists an integer $n$ which for all integers $k$ there exists another integer $d$ where $k+n=2d$’

Your interpretation is correct. Its negation says that there is some integer $k_n$ such that whatever integer $d$ is, $$k_n\ne2d-n.$$ Putting $k_n:=n+1$ shows that its negation is true.