%
\label{prop:zero-vacuous}
In $\tau$-Idx, zero is a passive, non-participating element:
\begin{enumerate}
    \item $\underline{0}$ is \textbf{not prime}:
          it fails the requirement $\underline{p} \geq \underline{2}$.
    \item $\underline{0}$ is \textbf{not a successor}:
          no $\underline{k} \in \tau$-Idx satisfies
          $\underline{k} + \underline{1} = \underline{0}$.
    \item $\underline{0}$ is \textbf{divisible by everything}:
          $\underline{a} \mid \underline{0}$ for all $\underline{a}$.
    \item $\underline{0}$ is the \textbf{unique multiplicative absorber}:
          if $\underline{a} \times \underline{n} = \underline{a}$
          for all $\underline{n}$,
          then $\underline{a} = \underline{0}$.
\end{enumerate}

(1) By definition
(Definition~\ref{def:internal-primes}):
$\underline{0} < \underline{2}$.

(2) If $\underline{k} + \underline{1} = \underline{0}$,
then by Proposition~\ref{prop:sum-zero-iff}
(Chapter~\ref{ch:swap-add-mul}),
$\underline{1} = \underline{0}$, a contradiction.

(3) For any $\underline{a}$:
$\underline{0} = \underline{a} \times \underline{0}$,
so $\underline{a} \mid \underline{0}$.

(4) Specialize to $\underline{n} = \underline{0}$:
$\underline{a} \times \underline{0} = \underline{a}$.
But $\underline{a} \times \underline{0} = \underline{0}$
(Proposition~\ref{prop:arithmetic-laws},
Chapter~\ref{ch:swap-add-mul}),
so $\underline{a} = \underline{0}$.