If $F({\boldsymbol{v}})={\boldsymbol{v}}'$ then for every encoding $M_{{\boldsymbol{v}}}$ of ${\boldsymbol{v}}$, there exists an encoding $M_{{\boldsymbol{v'}}}$ of ${\boldsymbol{v}}'$ such that $M_{{\boldsymbol{v}}}{\longrightarrow_{1}}{{\xrightarrow{*}_{\textit{Disc}}}}M_{{\boldsymbol{v}}'}$. Conversely, if $M_{{\boldsymbol{v}}}{\longrightarrow_{1}}{{\xrightarrow{*}_{\textit{Disc}}}}M_{{\boldsymbol{v}}'}$ for encodings $M_{{\boldsymbol{v}}}$ and $M_{{\boldsymbol{v'}}}$ of ${\boldsymbol{v}}$ and ${\boldsymbol{v}}'$ respectively, then $F({\boldsymbol{v}}) = {\boldsymbol{v'}}$.

For the first part, we construct a sequence $M_0 {{\xrightarrow{}_{\textit{Disc}}}} M_1 {{\xrightarrow{}_{\textit{Disc}}}} \dots {\xrightarrow{}_{\textit{Disc}}} M_{n-1} $ where $M_0 {{\overset{\text{\tinydef}}{=}}}(M_{{\boldsymbol{v}}}+1)$ and every step $M_{i-1} {{\xrightarrow{}_{\textit{Disc}}}} M_i$ adds tokens simulating the $i$th constraint of $F$. Since the TPN is non-consuming, we will have that $M_i\ge (M_{{\boldsymbol{v}}}+1)$, for all $i<n$. Consider now constraint $i'$, and assume w.l.o.g. that $i' = j \land k$ (the other case is analogous). There are two cases depending on ${\boldsymbol{v'}}[i]$.

1.  Case ${\boldsymbol{v'}}[i]=1$. By our assumption that $F({\boldsymbol{v}})={\boldsymbol{v'}}$ we know that ${\boldsymbol{v}}[j] =1$ and ${\boldsymbol{v}}[k] =1$. So $({\mathit{True}_{j}},1) \in (M_{{\boldsymbol{v}}}+1) \le M_{i-1}$ and $({\mathit{True}_{k}},1) \in (M_{{\boldsymbol{v}}}+1) \le M_{i-1}$. By construction of the net, there is a transition $i.B$ with ${\mathit{Pre}}(i.B) = {({\mathit{True}_{j}},1) + ({\mathit{True}_{k}},1)}$ and ${\mathit{Post}}(i.B) = {\mathit{Pre}}(i.B) + {({\mathit{True}_{i}},0)} $. This justifies step $M_{i-1} {\longrightarrow_{i.B}} M_{i}$ and therefore that $(True_i, 0) \in M_i\le M_{n-1}$. Also notice that no marking reachable from $M_0$ using only discrete steps can contain the token $({\mathit{False}_{i}},0)$. This is because these can only be produced by transitions 