1% Section: Discussion2\section{Discussion}\label{sec:discussion}34\paragraph{Competing explanations considered.} (1) \emph{A small-$n$ artifact.} A pattern seen for $n\le 30$ could fail later. This is excluded for $n\ge 26$ by proof (Lemma~\ref{lem:33} with the certified window), and the frozen holdout gave an honest out-of-sample test before the proof was complete; we note, however, that by the time the holdout values were computed the window $26\le n\le 154$ had been certified, so Lemma~\ref{lem:33} already guaranteed the prediction $12$ at every holdout $n$, and the holdout therefore tested the pipeline (witness replay, certificate generation and checking at new $n$) rather than the conjecture itself. (2) \emph{An artifact of the construction.} The constant $12$ could be a property of the $12$-outer-vertex construction while the true values are lower. The certified lower bounds exclude this, and the zero slack of Observation~\ref{obs:slack} shows that $12$ is the natural value of the boundary invariant. (3) \emph{A shared bug.} Agreement between solver paths could hide a common error (wrong adjacency, too short a round horizon, an over-constrained cardinality encoding). The two graph constructors, the three exhaustive and certificate paths with no shared code, the exhaustive validation of the encodings at small $n$, and the negative controls on the checker address this; the one bug that the campaign did find was a textual mis-description, not a computational error. (4) \emph{Residue dependence.} The inner structure depends on $\gcd(n,5)$, and $24\mid n$ was the only published family; neither affects the value for $n\ge 23$, and for $n\le 22$ the irregularity is explained by the isomorphisms to $k=3,4$ rather than by residues. (5) \emph{Prior containment.} The strongest risk is that \citet{alameda2018families} prove more than $\Z(\PP{24r}{5})=12$; we could not read the paper and claim nothing for $24\mid n$. The threshold statement and the values for $24\nmid n$ are not in the reported result.56\paragraph{Thresholds across $k$.} The sharp thresholds now known are $n_0=13$ for $k=3$ and $n_0=18$ for $k=4$ \citep{arxiv2610_24059}, and $n_0=23$ for $k=5$ (Theorem~\ref{thm:main}), with the value $2k+2$ in each case. The three thresholds are consistent with $n_0(k)=5k-2$, but three data points do not justify a conjecture, and the mechanisms differ: for $k=5$ the boundary method itself fails at $n=25=k^2$ because of the inner pentagons, and the pre-threshold values are governed by isomorphisms to smaller $k$. We record the question rather than a conjecture.78\begin{question}\label{q:threshold}9Is $\Z(\PP{n}{k})=2k+2$ for all $n\ge 5k-2$ and $k\ge 3$, with $\Z(\PP{5k-3}{k})<2k+2$? Does the first-force boundary method certify the lower bound for all $k$ with a parameter $p(k)$ and, for each $k$, which finitely many $n$ (such as $n=25$ for $k=5$) require a different argument?10\end{question}1112\paragraph{Maximum nullity.} Theorem~\ref{thm:main} gives $M(\PP{n}{5})\le 12$ for $n\ge 23$; \citet{alameda2018families} are reported to show equality for $24\mid n$ (as stated in \citealp[Remark~2.4]{rashidi2020petersen}; we could not read the primary text, see Table~\ref{tab:exact}). Whether $M(\PP{n}{5})=12$ for all $n\ge 23$ is open; we did not compute $M$.1314\paragraph{Methodological note.} The lower-bound method is borrowed; what the $k=5$ case adds is a clear view of its limits (Proposition~\ref{prop:p16}, Computational Result~\ref{cr:obstruction}, Observation~\ref{obs:slack}) and a certification discipline in which each machine-checked input is a DRAT proof for a formula whose relation to the theorem is a proved lemma, re-verified by a second program, with negative controls on the checker. We regard the explicit account of what is machine-checked and what is not (Appendix~\ref{sec:scope}) as part of the result.15