of a marking encoding, we have that $({\mathit{True}_{i}},0)\in M_{{\boldsymbol{v}}}$. By construction, there is a transition $i.B$ with ${\mathit{Pre}}(i.B) = {({\mathit{True}_{j}},1) + ({\mathit{True}_{k}},1)}$ and ${\mathit{Post}}(i.B) = {\mathit{Pre}}(i.B) + {({\mathit{True}_{i}},0)}$. By assumption, it holds that $(M_{{\boldsymbol{v}}}+1) {{\xrightarrow{*}_{\textit{Disc}}}} M_{{\boldsymbol{v}}}'$, where $M_{{\boldsymbol{v}}}{\longrightarrow_{1}} (M_{{\boldsymbol{v}}}+1)$. Note that $({\mathit{True}_{j}},1)\in (M_{{\boldsymbol{v}}}+1)$ and $({\mathit{True}_{k}},1)\in (M_{{\boldsymbol{v}}}+1)$. Hence, we have that ${\boldsymbol{v}}[j]=1$ and ${\boldsymbol{v}}[k]=1$, and therefore that $F({\boldsymbol{v}})[i] = {\boldsymbol{v'}}[i] = {\boldsymbol{v}}[j] \land {\boldsymbol{v}}[k]$.

2.  Case ${\boldsymbol{v'}}[i]=0$. Then $({\mathit{False}_{i}},0)\in M_{{\boldsymbol{v}}}$ and, since this token can only be produced by transitions $i.L$ or $i.R$, either $({\mathit{False}_{j}},1)\in (M_{{\boldsymbol{v}}}+1)$ or $({\mathit{False}_{k}},1)\in (M_{{\boldsymbol{v}}}+1)$.

    Therefore $({\mathit{False}_{j}},0)\in (M_{{\boldsymbol{v}}})$ or $({\mathit{False}_{k}},0)\in (M_{{\boldsymbol{v}}})$ and because $M_{{\boldsymbol{v}}}$ is an encoding of ${\boldsymbol{v}}$, this means that either ${\boldsymbol{v}}[j]=0$ or ${\boldsymbol{v}}[k]=0$. Therefore, $F({\boldsymbol{v'}})[i]= {\boldsymbol{v}}[j]\land {\boldsymbol{v}}[k] =0$.

\[thm:LB\] [$\exists$COVER ]{} is [$\mathsf{PSPACE}$]{}-hard for non-consuming TPN.

For a given monotone Boolean circuit, define a non-consuming TPN as above. By induction on $m\in{\mathbb{N}}$ using \[claim1\], we derive that there exists $m\in{\mathbb{N}}$ with $F^m({\boldsymbol{v}})={\boldsymbol{v}}'$ and ${\boldsymbol{v}}'[0]=1$ if, and only if, there exists encodings $M_{{\boldsymbol{v}}}$ of ${\boldsymbol{v}}$ and $M_{{\boldsymbol{v'}}}$ of ${\boldsymbol{v'}}$, with $M_{{\boldsymbol{v}}}{\xrightarrow{*}} M_{{\boldsymbol{v}}'}$. Moreover, if there is a marking $M$ such that $M_{{\boldsymbol{v}}}{\xrightarrow{*}} M$ and $0\in {{\it frac}(M)}$, where $M$ contains a token of age $0$, then $M\le M_{{\boldsymbol{v'}}}$ for some encoding $M_{{\boldsymbol{v}}'}$ of a vector 