% Section: Upper bounds \section{Upper bounds}\label{sec:upper} \begin{proposition}[the uniform bound $12$]\label{prop:P7} For every $n\ge 12$ the set $S=\{u_0,\dots,u_{11}\}$ is a zero-forcing set of $\PP{n}{5}$. Hence $\Z(\PP{n}{5})\le 12$ for all $n\ge 11$. \end{proposition} \begin{proof} For $n=12$, $S$ is the whole outer layer and each $u_i$ forces $v_i$ through its spoke. Let $n\ge 13$. \emph{Phase 1.} For $i=1,\dots,10$ perform $u_i\to v_i$: the other neighbours $u_{i-1},u_{i+1}$ of $u_i$ lie in $S$ because $0\le i-1$ and $i+1\le 11$. Then $v_5\to v_0$ ($N(v_5)=\{u_5,v_{10},v_0\}$, the first two black) and $v_6\to v_{11}$ ($N(v_6)=\{u_6,v_1,v_{11}\}$). Now columns $0,\dots,11$ are black and columns $12,\dots,n-1$ white (column $n-1\ne 11$ since $n\ge 13$). \emph{Phase 2.} For $j=12,\dots,n-1$ in increasing order perform $u_{j-1}\to u_j$ and then $v_{j-5}\to v_j$. Invariant before stage $j$: columns $