%
\label{prop:no-second-linearity}
The ABCD coordinate chart $\Phi$ is not linear:
there exist objects $X, Y \in \tau\text{-Idx}$
such that
\[
    \Phi(X + Y) \;\neq\; \Phi(X) + \Phi(Y),
\]
where $+$ denotes $\alpha$-ray addition
(iterated $\rho$) on both sides,
applied componentwise on the right.

[Proof sketch]
Take $X = \underline{2}$ and $Y = \underline{3}$.
Then $X + Y = \underline{5}$.
The ABCD coordinates are:
\begin{align*}
    \Phi(\underline{2})
    &= (\underline{2},\, \underline{1},\, \underline{1},\, \underline{1}), \\
    \Phi(\underline{3})
    &= (\underline{3},\, \underline{1},\, \underline{1},\, \underline{1}), \\
    \Phi(\underline{5})
    &= (\underline{5},\, \underline{1},\, \underline{1},\, \underline{1}).
\end{align*}
Componentwise addition of the first two tuples gives
$A = \underline{2} + \underline{3} = \underline{5}$,
which coincidentally matches.
But now take $X = \underline{2}$ and $Y = \underline{4}$,
so $X + Y = \underline{6}$.
We have:
\begin{align*}
    \Phi(\underline{4})
    &= (\underline{2},\, \underline{2},\, \underline{1},\, \underline{1}), \\
    \Phi(\underline{6})
    &= (\underline{3},\, \underline{1},\, \underline{1},\, \underline{2}).
\end{align*}
Componentwise addition gives
$(A, B, C, D) = (\underline{2} + \underline{2},\,
\underline{1} + \underline{2},\,
\underline{1} + \underline{1},\,
\underline{1} + \underline{1})
= (\underline{4},\, \underline{3},\, \underline{2},\, \underline{2})$,
but $\Phi(\underline{6})
= (\underline{3},\, \underline{1},\, \underline{1},\, \underline{2})$.
These differ in all of $A$, $B$, and $C$.

The structural reason is that the NF encoding interleaves
multiplicative data ($A$, $B$) and exponential data ($C$)
through the greedy peel
(Definition~\ref{def:genealogical-decomposition}).
Addition on the $\alpha$-ray can change
the \emph{prime factorization} of the sum,
redistributing mass across all four coordinates
in a nonlinear fashion.