(2); (1) – (3); (1) – (4); (2) – (5); (2) – (6); (3) – (5); (3) – (7); (4) – (6); (4) – (7); (5) – (8); (6) – (8); (7) – (8);

$\operatorname{Sign}$ induces the following best correct approximation of ${\ensuremath{\mathit{sq}}}$: $$\begin{aligned}
{\ensuremath{\mathit{sq}}}^{\operatorname{Sign}} = \{&\varnothing \mapsto \varnothing, \mathbb{Z}_{<0} \mapsto 
\mathbb{Z}_{> 0}, 0 \mapsto 0, \mathbb{Z}_{> 0} \mapsto
\mathbb{Z}_{> 0}, \mathbb{Z}_{\leq 0} \mapsto \mathbb{Z}_{\geq 0}, \\
& \mathbb{Z}_{\neq 0} \mapsto \mathbb{Z}_{>0},
\mathbb{Z}_{\geq 0} \mapsto \mathbb{Z}_{\geq 0},
\mathbb{Z} \mapsto \mathbb{Z}_{\geq 0}\}.\end{aligned}$$ Let us characterize the correctness kernel $\mathscr{K}_{{\ensuremath{\mathit{sq}}}}(\operatorname{Sign})$ by Theorem \[kernel\]. We have that $\operatorname{img}({\ensuremath{\mathit{sq}}}^{\operatorname{Sign}})= \{\varnothing, \mathbb{Z}_{>0}, 0, 
\mathbb{Z}_{\geq 0}\}$. Moreover, $$\begin{aligned}
\max (\{x\in \operatorname{Sign}~|~ {\ensuremath{\mathit{sq}}}^{\operatorname{Sign}}(x) = \varnothing\}) &= \{\varnothing\}\\
\max (\{x\in \operatorname{Sign}~|~ {\ensuremath{\mathit{sq}}}^{\operatorname{Sign}}(x) = \mathbb{Z}_{>0}\}) &= \{\mathbb{Z}_{\neq 0}\}\\ 
\max (\{x\in \operatorname{Sign}~|~ {\ensuremath{\mathit{sq}}}^{\operatorname{Sign}}(x) = 0\}) &= \{0\}\\
\max (\{x\in \operatorname{Sign}~|~ {\ensuremath{\mathit{sq}}}^{\operatorname{Sign}}(x) = \mathbb{Z}_{\geq 0}\}) &= \{\mathbb{Z}\}\end{aligned}$$ Therefore, $\bigcup_{y\in \operatorname{img}({\ensuremath{\mathit{sq}}}^{\operatorname{Sign}})} \max (\{x\in \operatorname{Sign}~|~ {\ensuremath{\mathit{sq}}}^{\operatorname{Sign}}(x) = y\}) 
= \{\varnothing,\mathbb{Z}_{\neq 0},0,\mathbb{Z}\}$ so that, by Theorem \[kernel\]: $$\mathscr{K}_{{\ensuremath{\mathit{sq}}}}(\operatorname{Sign}) = \operatorname{Cl}_{\cap}(\{\varnothing, \mathbb{Z}_{> 0}, 
0,\mathbb{Z}_{\geq 0}, \mathbb{Z}_{\neq 0}, \mathbb{Z}\}) = \operatorname{Sign}\smallsetminus
\{\mathbb{Z}_{< 0}, \mathbb{Z}_{\leq 0}\}.$$ Thus, it turns out that we can safely remove the abstract values $\mathbb{Z}_{< 0}$ and $\mathbb{Z}_{\leq 0}$ from $\operatorname{Sign}$ and still preserve the same b.c.a. as $\operatorname{Sign}$ does. Besides, we cannot 