sets closed under wirings, which turns out to be a difficult task [@Lang_2014].

Considering two copies of an initial box $p(ab\vert xy)$ given by , after the wiring we obtain $$\begin{aligned}
\nonumber
    W(p,p)&=c_0^2 B_0^0+c_0\sum_{i=1}^{8}c_{i}(B^0_{i}+B_0^{i})\\
    &+\sum_{i,j=1}^{8}c_{i}c_{j}B^{i}_{j},\end{aligned}$$ where $$\begin{aligned}
\label{Bs}
 B_0^0(ab|xy)= & & W(PR(ab|xy),PR(ab|xy)) \\ \nonumber
    B_{i}^0(ab|xy)= & & W(PR(ab|xy),L_{i}(ab|xy)) \\ \nonumber
    B_0^{i}(ab|xy)= & &W(L_{i}(ab|xy),PR(ab|xy)) \\ \nonumber
    B^{i}_{j}(ab|xy)= & &W(L_{i}(ab|xy),L_{j}(ab|xy))\end{aligned}$$

The nonlocality of the initial distribution is $NL(p)=c_0$, since $NL$ is linear for the distribution and $NL(PR)=1$ and $NL(L_{i})=0$ for all $i$. For the wired distribution we obtain directly an upper bound for its nonlocality, given by $$\begin{aligned}
\label{NLwired}
    & &NL(W(p,p)) \leq c_0^2 NL(B^0_0) \\ \nonumber & & +c_0\sum_{i=1}^{8}c_{i}(NL(B^0_{i})+NL(B_0^{i})).\end{aligned}$$ Clearly, to have nonlocality distillation we need $NL(W(p,p)) > NL(p)$ thus implying that $$\begin{aligned}
& & c_0NL(B^0_0) \\ \nonumber 
& & +\sum_{i=1}^{8}c_{i}(NL(B^0_{i})+NL(B_0^{i})) > 1. \end{aligned}$$

Since $c_0+\sum_{i=1}^{8}c_{i}=1$, the condition for distillation becomes $$c_0(NL(B_0 ^0)-1)+\sum_{i=1}^{8}c_{i}(NL(B^0_{i})+NL(B^{i}_0)-1)>0$$ As hinted by the expression above, looking at the nonlocality of the wired terms in can provide us with the necessary information to decide whether a given correlation is distilablle or not, something we will explore in what follows.

By testing all $82^4$ possible deterministic wirings on the wired distributions $B^0_0$, $B^{i}_0$ and $B_{i}^0$ one can prove that $NL(B^0_0)=\left\{0,1/2,1 