Let $K$ be a field and let
\[
  \Omega=\{+,-,\times,/\}
\]
be the ordinary binary arithmetic signature.  Division is partial in the
affine chart: a denominator must be nonzero at the moment at which the
operation is performed.

An arithmetic expression is a planar binary tree.  When its internal vertices
form a single chain, its operations acquire a unique temporal order, and the
expression can be written in two extreme ways: the already-computed subtree may
occupy the first operand slot or the second.  This section fixes that
convention, proves the elementary facts on which every later chapter depends,
and isolates the one algebraic law that relates the two expansions of a common
expression.

\subsection{Chronological comb conventions}

\begin{definition}[Left- and right-expanded one-hole expressions]
\label{def:p0-left-right-combs}
Fix constants $c_1,\ldots,c_n\in K$ and operation labels
$\omega_1,\ldots,\omega_n\in\Omega$.  Starting from the hole $z$, define
recursively
\begin{equation}\label{eq:p0-comb-recursions}
  L_0[z]=z,
  \qquad
  L_k[z]=\omega_k(L_{k-1}[z],c_k),
\end{equation}
\begin{equation}\label{eq:p0-right-comb-recursions}
  R_0[z]=z,
  \qquad
  R_k[z]=\omega_k(c_k,R_{k-1}[z]).
\end{equation}
The expression $L_n$ is a \emph{left-expanded comb}; $R_n$ is a
\emph{right-expanded comb}.  The index $k$ is the evaluation time: the
innermost operation has index $1$ and is evaluated first.
These recursions first define syntax.  Ordinary evaluation is a partial
interpretation of that syntax: a history is admissible at an input only if
every operation is defined at its actual operands, including all intermediate
denominators.
\end{definition}

\begin{convention}[Scope of the words ``left'' and ``right'']
\label{conv:p0-left-right-convention}
Throughout this paper, ``left-expanded'' and ``right-expanded'' refer only to
the recursion \eqref{eq:p0-comb-recursions}--\eqref{eq:p0-right-comb-recursions}:
the active subtree occupies operand slot $(1)$ in the first case and operand
slot $(2)$ in the second.  The index is chronological.  Every later use of the
words must be reducible to \eqref{eq:p0-comb-recursions}--\eqref{eq:p0-right-comb-recursions};
the planar direction in which a tree happens to be drawn is not part of the
convention.
\end{convention}

Thus, in fully parenthesized form,
\[
  L_n[z]
  =\omega_n(\cdots\omega_2(\omega_1(z,c_1),c_2)\cdots,c_n),
\]
whereas
\[
  R_n[z]
  =\omega_n(c_n,\omega_{n-1}(c_{n-1},\cdots\omega_1(c_1,z)\cdots)).
\]
In the first expression the active subtree is the left child at every internal
vertex.  In the second it is the right child.

\begin{figure}[t]
\centering
\begin{tikzpicture}[
  op/.style={circle,draw=black!65,fill=blue!7,minimum size=7mm,inner sep=0pt},
  leaf/.style={rectangle,draw=black!45,fill=gray!5,minimum height=6mm,
              minimum width=8mm,inner sep=1pt},
  every node/.style={font=\small}
]
  \node[font=\bfseries] at (-3.2,2.65) {left-expanded};
  \node[op] (l3) at (-3.2,2.0) {$\omega_3$};
  \node[op] (l2) at (-4.0,1.15) {$\omega_2$};
  \node[leaf] (lc3) at (-2.35,1.15) {$c_3$};
  \node[op] (l1) at (-4.75,0.3) {$\omega_1$};
  \node[leaf] (lc2) at (-3.25,0.3) {$c_2$};
  \node[leaf] (lz) at (-5.35,-0.55) {$z$};
  \node[leaf] (lc1) at (-4.15,-0.55) {$c_1$};
  \draw (l3)--(l2); \draw (l3)--(lc3);
  \draw (l2)--(l1); \draw (l2)--(lc2);
  \draw (l1)--(lz); \draw (l1)--(lc1);
  \draw[-{Latex[length=2mm]},very thick,blue!60!black]
    (-5.75,-0.9) .. controls (-5.2,-1.25) and (-3.6,-1.25) .. (-2.8,-0.9)
    node[midway,below,font=\scriptsize] {evaluation follows the left spine outward};

  \node[font=\bfseries] at (3.2,2.65) {right-expanded};
  \node[op] (r3) at (3.2,2.0) {$\omega_3$};
  \node[leaf] (rc3) at (2.35,1.15) {$c_3$};
  \node[op] (r2) at (4.0,1.15) {$\omega_2$};
  \node[leaf] (rc2) at (3.25,0.3) {$c_2$};
  \node[op] (r1) at (4.75,0.3) {$\omega_1$};
  \node[leaf] (rc1) at (4.15,-0.55) {$c_1$};
  \node[leaf] (rz) at (5.35,-0.55) {$z$};
  \draw (r3)--(rc3); \draw (r3)--(r2);
  \draw (r2)--(rc2); \draw (r2)--(r1);
  \draw (r1)--(rc1); \draw (r1)--(rz);
  \draw[-{Latex[length=2mm]},very thick,orange!75!black]
    (5.75,-0.9) .. controls (5.2,-1.25) and (3.6,-1.25) .. (2.8,-0.9)
    node[midway,below,font=\scriptsize] {evaluation follows the right spine outward};
\end{tikzpicture}
\caption{The two pure combs under the chronological indexing convention.
Both have one chain of internal dependencies; the active subtree occupies
opposite operand slots.}
\label{fig:p0-two-combs}
\end{figure}

\begin{proposition}[Unique internal order]
\label{prop:p0-combs-unique-order}
Every finite left- or right-expanded comb has a unique legal order of internal
evaluation, namely the chronological order $1,2,\ldots,n$.
\end{proposition}

\begin{proof}
At stage $k$, the only internal child of the $k$th vertex is the subtree formed
at stage $k-1$; the other child is the leaf $c_k$.  Hence the internal vertices
form one dependency chain
\[
  v_1<v_2<\cdots<v_n.
\]
A finite chain has a unique linear extension.
\end{proof}

This proposition is deliberately weaker than the full sequential-tree
classification of Paper~I.  Here the combs are given by explicit grammars;
Paper~I proves intrinsically that every binary expression tree with a unique
internal evaluation order has one such spine after operand slots are recorded.

\subsection{One-hole sections and opposite operations}

For $\omega\in\Omega$ and $c\in K$, write
\[
  C^{(1)}_{\omega,c}[z]=\omega(z,c),
  \qquad
  C^{(2)}_{\omega,c}[z]=\omega(c,z).
\]
Then $L_n$ is a chronological word of pure slot-$(1)$ sections and $R_n$ a
word of pure slot-$(2)$ sections.

Define the opposite binary operation by
\[
  \omega^{\mathrm{op}}(u,v):=\omega(v,u).
\]

\begin{proposition}[Comb opposition]
\label{prop:p0-comb-opposition}
With chronological indexing, a right-expanded comb for operations
$\omega_1,\ldots,\omega_n$ is exactly a left-expanded comb for the opposite
operations:
\[
  R_k[z]
  =\omega_k^{\mathrm{op}}(R_{k-1}[z],c_k).
\]
Consequently, the two comb grammars are isomorphic after replacing every
operation by its opposite and exchanging the operand slots.  Their planar
trees retain those slots.  For commutative operations with symmetric domains,
their partial operator interpretations agree; this does not identify the
planar trees or the marked histories.
\end{proposition}

\begin{proof}
By definition,
\[
  \omega_k(c_k,R_{k-1}[z])
  =\omega_k^{\mathrm{op}}(R_{k-1}[z],c_k).
\]
Apply this identity at every stage.  If $\omega_k$ is commutative, then
$\omega_k^{\mathrm{op}}=\omega_k$ as partial operations, including their
domains.  Interchanging the slots is the corresponding mirror of the
planar comb.  It changes syntax even when it preserves the interpreted map.
\end{proof}

The result is combinatorial, not an assertion that the two arithmetic maps are
always equal.  Their elementary one-hole maps are:

\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{lll}
\toprule
Operation & active value in slot $(1)$ & active value in slot $(2)$ \\
\midrule
addition & $z+c$ & $c+z=z+c$ \\
subtraction & $z-c$ & $c-z$ \\
multiplication & $cz$ & $cz$ \\
division & $z/c$ & $c/z$ \\
\bottomrule
\end{tabular}
\end{center}

The first-slot division $z/c$ is defined on all of $K$ if $c\ne0$ and
nowhere if $c=0$.  Second-slot division $c/z$ has ordinary domain
$K\setminus\{0\}$ for every $c$, including $c=0$: cancelling a zero
numerator does not restore the missing input.  Every division guard belongs
to the original history.  A later simplification or multiplication by zero
cannot make an earlier undefined division admissible.

All first-slot maps other than division are affine self-maps; first-slot
division is affine when its constant denominator is nonzero.  An affine
self-map $az+b$ belongs to the affine \emph{group} only when $a\ne0$.
Thus multiplication by zero is a permitted arithmetic expression and a
constant affine self-map, but not an invertible affine motion.  Second-slot
subtraction has multiplier $-1\ne0$ in any field.  For $c\ne0$,
second-slot division has an invertible projective continuation with pole
$0$; it sends $0$ to $\infty$ and $\infty$ to $0$.  When $c=0$, the
ordinary partial map is zero on its domain and has no invertible
M\"obius continuation.  The projective statement does not declare an
ordinary division by zero admissible.

\begin{corollary}[Where the algebraic symmetry breaks]
\label{cor:p0-affine-projective-break}
If the operation alphabet is restricted to $+$ and $\times$, pure left and
pure right expansion have the same one-hole operator algebra of affine
self-maps.  Restricting multiplicative constants to be nonzero gives affine
group elements; addition and subtraction preserve this group boundary.
First-slot division with nonzero constant denominator also remains inside
that group.  Once second-slot division $c/z$ with $c\ne0$ is admitted
and projectively continued, the right-expanded language leaves the affine
stabilizer of infinity.  It contains inversion itself when $c=1$ is allowed.
\end{corollary}

\begin{proof}
The first two assertions follow from the table.  For $c\ne0$,
\[
  \frac{c}{z}=D_c\circ J(z),
  \qquad D_c(z)=cz,\quad J(z)=\frac1z.
\]
The map $J$ exchanges $0$ and $\infty$, so it does not stabilize infinity.
In homogeneous coordinates the same transformation is
\[
  [X:Y]\longmapsto[cY:X],
  \qquad
  \begin{pmatrix}0&c\\1&0\end{pmatrix},
  \qquad \det=-c\ne0.
\]
For $c=0$, the representative $(0,1)$ is sent to $(0,0)$, so no
projective transformation is defined by this matrix.  The nonzero condition
is therefore necessary.  The non-affine assertion concerns this projective
transformation, not interpolation of its values on a finite subset of $K$.
\end{proof}

\subsection{The distributive law as an expansion rewrite}
\label{subsec:p0-distributivity}

Two different expansions may induce the same arithmetic operator.  The law used
here to relate two such expansions is distributivity, and it is worth isolating
because every later chapter compares expansions that share an operator or a
value.  Other ring identities (associativity, the unit laws) also relate
expansions with a common operator; what is special about distributivity is that
it moves an additive step across a multiplicative one, which is why it, and not
the others, produces the charge transport computed in
Example~\ref{ex:p0-distributive-charge-transport}.

\begin{definition}[Distributive expansion]
\label{def:p0-distributive-expansion}
Let $p,k\in K$.  The \emph{distributive expansion} of the two-step slot-$(1)$
word
\begin{equation}\label{eq:p0-distributive-expansion}
  z\;\longmapsto\;k(z+p)=(kz+kp)
\end{equation}
is the slot-$(1)$ word obtained by reading the right-hand side as a multiplication
followed by a translation:
\[
  z\;\longmapsto\;kz\;\longmapsto\;kz+kp .
\]
The two words have the same length, the same operation alphabet, and the same
induced operator $z\mapsto kz+kp$ on $K$.  Their marked chronological data
are respectively
\[
  ((+,1,p),(\times,1,k)),
  \qquad
  ((\times,1,k),(+,1,kp)),
\]
where the middle entry records the active operand slot.  Forgetting operation
labels and slots gives the \emph{chronological scalar projection}
$(p,k)$ and $(k,kp)$.  This projection is not a complete marked history;
it also differs from grouping constants by their additive or multiplicative
role.  A zero multiplicative constant is allowed here as arithmetic data,
without making the associated constant map an affine group element.
\end{definition}

\begin{proposition}[Distributivity is operator-preserving and expansion-changing]
\label{prop:p0-distributive-operator-identity}
For all $p,k\in K$ the induced operators of the two words in
\eqref{eq:p0-distributive-expansion} agree:
\begin{equation}\label{eq:p0-distributive-identity}
  \bigl(z\mapsto z+p\bigr)\ \text{followed by}\ \bigl(z\mapsto kz\bigr)
  \;=\;
  \bigl(z\mapsto kz\bigr)\ \text{followed by}\ \bigl(z\mapsto z+kp\bigr).
\end{equation}
The two marked words are distinct for every $p,k$, since their first
operation labels differ.  Their chronological scalar projections coincide
precisely when
\[
  (p,k)=(0,0)\quad\text{or}\quad(p,k)=(1,1).
\]
In particular, if $p\ne0$ and $k\ne1$, then the two expansions share their
operator and differ even after projecting to chronological scalar data.
\end{proposition}

\begin{proof}
Both sides send $z$ to $kz+kp$, which is the distributive law in $K$.
Marked-word equality would require $+=\times$ at the first position in the
operation signature, which is impossible, independently of any accidental
equality of interpreted maps over a particular field.  Equality of the
scalar projections requires $p=k$ and $k=kp$.  If $k=0$, the first condition
gives $p=0$.  If $k\ne0$, cancellation in $k\cdot1=k\cdot p$ gives
$p=1$, and then $k=1$.  Conversely both listed pairs satisfy the two
equations.  The final assertion excludes both cases.  Thus operator
equality, scalar-projection equality and marked-history equality are
separate judgments.
\end{proof}

\begin{example}[Distributivity transports the additive charge]
\label{ex:p0-distributive-charge-transport}
Take $K=\mathbb R$, $p=1$, $k=e^{q}$ with $q\ne0$, and compare the two words
of Proposition~\ref{prop:p0-distributive-operator-identity}.  Both induce
$z\mapsto e^{q}z+e^{q}$.  Their accumulated charges, in the sense fixed in
Section~\ref{sec:p0-acs-tearing}, are
\[
  (A,M)=(1,q)
  \qquad\text{and}\qquad
  (A,M)=(e^{q},q).
\]
The multiplicative charge is the same because the two words differ only in
where the additive step is performed; the additive charge differs by exactly
the scale $e^{q}$ that separates the two positions.  In the notation of
Definition~\ref{def:acs}, the weighted integrals agree,
\[
  \int_{C_{\gamma}}e^{-M}\dd A
  =1\cdot e^{0}=1,
  \qquad
  \int_{C_{\delta}}e^{-M}\dd A
  =e^{q}\cdot e^{-q}=1,
\]
so that Proposition~\ref{prop:acs-evaluation} returns the common value
$e^{q}(x+1)$ along both charge paths.
\end{example}

The two histories of Example~\ref{ex:p0-distributive-charge-transport} share
their motion and differ in their charge.  The difference is an instance of the
transport elements of Theorem~\ref{thm:p0-descent}: the chronological word of
$\gamma$ followed by the inverse of $\delta$ evaluates to the identity motion
while carrying additive charge $p(1-e^{q})$, which is the general phenomenon
that section names and measures.

Example~\ref{ex:p0-distributive-charge-transport} is the elementary core of the
later discussion of tearing.  Distributivity does not merely reorganize
symbols: it moves additive charge through multiplicative charge, and the moved
amount is exactly the scale accumulated in between.  An algebraic identity of
operators therefore becomes a statement about two different charge histories.

\begin{remark}[Where expansion stops]
\label{rem:p0-expansion-boundary}
Distributivity is a law of the ring $(\times,+)$ and not a universal law of the
whole signature $\Omega$.  Its analogues for the two remaining operations
behave differently, and the difference is instructive.

\begin{enumerate}
  \item For slot-$(2)$ subtraction,
  \[
    (c-z)k=kc-kz=(-k)z+kc .
  \]
  This is the distributive law again, and the transported constant is $kc$;
  the resulting one-hole map is still affine, but its multiplier is $-k$.
  Expansion therefore remains inside the affine self-maps; its map belongs
  to the affine group when $k\ne0$.

  \item For slot-$(2)$ division,
  \[
    \frac{c}{z}\,k=\frac{kc}{z},
    \qquad z\ne0 .
  \]
  This is an identity of partial functions, not an expansion: the right-hand
  side has the original excluded input $z=0$, even if $kc=0$.  If $kc\ne0$,
  its projective continuation is inversion followed by an invertible scaling,
  outside the affine group.  If $kc=0$, it is a constant partial map with
  the same excluded input, not an invertible projective generator.  It is
  already an ordinary slot-$(2)$ comb; what requires a generator outside the
  affine group is the non-degenerate projective continuation.
\end{enumerate}

Consequently the invertible expansion calculus of this section is an affine
calculus, and the non-degenerate projective step $c/z$, $c\ne0$, is
where that group boundary ends.  Section~\ref{sec:p0-ripples}
reads that step contravariantly instead, as a condition on outputs rather than
a motion of the value.
\end{remark}

\subsection{Expansion and mechanical computation}

The distributive law is what makes mechanical expansion possible at all.  A
symbolic engine that reduces a polynomial expression to a normal form does so
by distributing products over sums and collecting like terms; this is the
algebraic core of the mechanization of mathematics, whose systematic programme
was set out by Wu Wen-tsun \cite{Wu2000MathematicsMechanization}.  In that
tradition expansion is a means: the process is eliminated in favour of a
canonical object, and the mechanical content is transferred to an algorithm
acting on that object.

The present paper keeps the expansions.  What
Example~\ref{ex:p0-distributive-charge-transport} shows is that the expansions
which distributivity relates are distinguishable objects: they carry different
one-hole words, different intermediate values, and different charges, while
sharing one operator and one endpoint.  This is the sense in which AEG studies
process rather than result.  The value of an expression is one coordinate of
its expansions, not the object itself; the geometry developed in the following
sections is the geometry of the expansions.

\subsection{Point propagation and condition propagation}

The same one-hole word supports two elementary readings.  Given maps
$F_1,\ldots,F_n$ and a point $z_0$, the prefix operators
\[
  G_k=F_k\circ\cdots\circ F_1
\]
push the point forward:
\[
  z_k=G_k(z_0).
\]
This is the \emph{point propagation} used for paths.
For ordinary partial maps, put $D_0=K$ and $G_0=\id$, and recursively set
\[
  D_k=\{z\in D_{k-1}:G_{k-1}(z)\in\operatorname{dom}F_k\}.
\]
Then $G_k$ is evaluated only on $D_k$.  This prefix-domain rule retains
every original division condition even when a simplified final formula has
a larger natural domain.

Alternatively, given a family of output level sets $\{\Sigma_t\}$, the same
prefix operator pulls each level set back:
\[
  \Sigma_{k,t}=G_k^{-1}(\Sigma_t).
\]
For partial maps this means
$\Sigma_{k,t}=\{z\in D_k:G_k(z)\in\Sigma_t\}$; it is an inverse
image, not a claim that $G_k$ has an inverse function.
This is \emph{condition propagation}.  It is contravariant: an outer condition
is transported toward the input by inverse image.  Sections~\ref{sec:p0-aes}
and~\ref{sec:p0-ripples} show that the first reading yields a path in a space of
arithmetic motions, while the second yields what we shall call ripple geometry.