% =====================================================================
%  Sections 2-6: local structure, angle identities, convex dissections
%  and chords, boundary pieces, pinwheel skeletons, and the complete
%  classification for k = 3 and k = 4 pieces.
%  Insertable into the A389392 manuscript: it only needs the macros and
%  theorem environments defined in the preamble of nota_completitud.tex
%  and the notation of Section 1 (T, n, A, B, C, \trap{b}{s}, G).
% =====================================================================

\section{Local structure and two angle identities}\label{sec:local}

For a vertex $v$ of $G$ let $a(v)$ (resp.\ $c(v)$) be the number of pieces
having a corner at $v$ with angle $60^\circ$ (resp.\ $120^\circ$), and let
$h(v)$ be the number of pieces containing $v$ in the relative interior of one
of their sides. Since every piece has two corners of each kind,
\begin{equation}\label{eq:ac}
  \sum_{v} a(v)=\sum_{v} c(v)=2k .
\end{equation}

\begin{lemma}[local structure]\label{lem:local}
\begin{enumerate}[label=\textup{(\alph*)}]
\item At a corner of $T$ we have $(a,c,h)=(1,0,0)$.
\item At a boundary vertex we have $h=0$ and $(a,c)\in\{(1,1),(3,0)\}$; we say
  that the vertex is of type $B_1$ or $B_3$. A $B_1$ vertex is incident with
  exactly one interior edge; a $B_3$ vertex is incident with exactly two, which
  form angles of $60^\circ$ with the side of $T$ and with each other.
\item At an interior vertex we have $h\le 1$ and $a+2c+3h=6$. Hence $(a,c,h)$
  is one of $(1,1,1)$, $(3,0,1)$, $(0,3,0)$, $(2,2,0)$, $(4,1,0)$, $(6,0,0)$,
  and we say that the vertex is of type $\mathrm{T}$, $\mathrm{T}'$,
  $\mathrm{Y}$, $\mathrm{X}$, $\mathrm{W}$, $\mathrm{Z}$, respectively. The
  degrees of these vertices are $3,4,3,4,5,6$.
\end{enumerate}
At a vertex $Q$ of type $\mathrm{T}$ two of the three edges are collinear; the
straight segment they form is the \emph{line} of $Q$, and the third edge is the
\emph{stem} of $Q$. The piece containing $Q$ in the relative interior of a side
lies on the side of the line opposite to the stem. At a vertex of type
$\mathrm{Y}$ any two of the three edges form an angle of $120^\circ$.
\end{lemma}

\begin{proof}
The pieces containing $v$ cover a neighbourhood of $v$ in $T$. Each of them
contributes an angle of $60^\circ$ or $120^\circ$ (if $v$ is one of its
corners) or of $180^\circ$ (if $v$ lies in the relative interior of one of its
sides), and these contributions add up to $60^\circ$, $180^\circ$ or
$360^\circ$ in cases (a), (b), (c), respectively. In (a) an angle of $60^\circ$
cannot be split. In (b), a piece containing $v$ in the relative interior of a
side has this side on $\partial T$, because the piece lies in $T$; it then
covers a half-disc around $v$, so no piece has a corner at $v$, and $v$ would
not be a vertex. In (c), two such pieces would have their sides on a common
line and lie on opposite sides of it, and again no piece would have a corner at
$v$. The remaining assertions are arithmetic: an interior vertex has degree
$a+c+h$, the number of pieces around it, and a boundary vertex has degree
$a+c+1$; at a vertex of type $\mathrm{T}$ the piece with angle $180^\circ$
determines the line, and the stem separates the two pieces with angles
$60^\circ$ and $120^\circ$.
\end{proof}

We write $B_1,B_3,t,t',y,x,w,z$ for the numbers of vertices of the respective
types, and $B=B_1+B_3$ for the number of boundary vertices.

\begin{lemma}[angle identities]\label{lem:identities}
For every $k$-tiling of $T$,
\begin{align}
  y&=1+B_3+t'+w+2z, \label{eq:I1}\\
  B_1+3B_3+t+3t'+2x+4w+6z&=2k-3. \label{eq:I2}
\end{align}
In particular, every tiling has a vertex of type $\mathrm{Y}$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:local}, $a(v)-c(v)$ equals $1$ at the corners of $T$; $0$ at
vertices of types $B_1$, $\mathrm{T}$, $\mathrm{X}$; $3$ at vertices of types
$B_3$, $\mathrm{T}'$, $\mathrm{W}$; $6$ at vertices of type $\mathrm{Z}$; and
$-3$ at vertices of type $\mathrm{Y}$. By \eqref{eq:ac} the sum of all these
numbers is $0$, which is \eqref{eq:I1}. Next we add up all angles at all
vertices in two ways. Piece by piece, each piece contributes $360^\circ$ at its
corners and $180^\circ$ for every vertex lying in the relative interior of one
of its sides, so the total is $360k+180(t+t')$ degrees. Vertex by vertex, the
total is $3\cdot 60+180B+360(t+t'+y+x+w+z)$ degrees. Hence
\[
  2k=1+B_1+B_3+t+t'+2(y+x+w+z),
\]
and eliminating $y$ by means of \eqref{eq:I1} gives \eqref{eq:I2}.
\end{proof}

\section{Dissections of convex polygons, full strips and chords}\label{sec:chords}

\begin{lemma}\label{lem:twoconvex}
If a convex polygon $P$ is the union of two convex polygons $P_1,P_2$ with
disjoint interiors, then there is a line $\ell$ such that $P_1$ and $P_2$ are
the intersections of $P$ with the two closed half-planes bounded by $\ell$.
\end{lemma}

\begin{proof}
By the separation theorem there is a line $\ell$ with $P_i\subseteq H_i$, where
$H_1,H_2$ are the closed half-planes bounded by $\ell$. A point of $P$ in the
interior of $H_1$ does not lie in $P_2$, hence lies in $P_1$; taking closures,
$P\cap H_1\subseteq P_1\subseteq P\cap H_1$. The same argument applies to
$P_2$.
\end{proof}

\begin{lemma}[lattice chords]\label{lem:chords}
Let $P$ be a convex polygon with sides on lattice lines, and let $\sigma$ be a
lattice segment joining two points of $\partial P$ and meeting the interior of
$P$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item If $P$ is an equilateral triangle, then $\sigma$ is parallel to a side and
  cuts off an equilateral triangle.
\item If $P=\trap{b}{s}$, then exactly one of the following holds:
  \textup{(i)} $\sigma$ is parallel to the bases and joins the two legs, and
  both parts are trapezoids;
  \textup{(ii)} $\sigma$ is parallel to a leg $\lambda$ and joins the two bases,
  the part adjacent to $\lambda$ is a parallelogram and the other part is a
  trapezoid;
  \textup{(iii)} $\sigma$ cuts off an equilateral triangle at one of the two
  $60^\circ$ corners.
\item If $P$ is a parallelogram, then either $\sigma$ is parallel to two sides
  of $P$ and both parts are parallelograms, or $\sigma$ lies in the third
  lattice direction and either joins two opposite sides (both parts are then
  trapezoids) or cuts off an equilateral triangle at a $60^\circ$ corner.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The endpoints of $\sigma$ lie on two different sides, so $\sigma$ is
parallel to the third side. (b) After a symmetry of the lattice,
$P=\{0\le y\le s,\ x\ge 0,\ x+y\le b\}$. A lattice segment through an interior
point lies on a line $y=c$, $x=c$ or $x+y=c$. A segment on $y=c$ joins the
legs, as in (i). A segment on $x=c$ joins $y=0$ to the short base $y=s$ if
$c\le b-s$, which is case (ii) with the parallelogram $\{0\le x\le c\}$;
otherwise it joins $y=0$ to the leg $x+y=b$ and cuts off a triangle at $(b,0)$.
Segments on $x+y=c$ are treated symmetrically. (c) After a symmetry,
$P=\{0\le x\le p,\ 0\le y\le q\}$ with $60^\circ$ corners at $(0,0)$ and
$(p,q)$; segments on $x=c$ or $y=c$ give two parallelograms, and a segment on
$x+y=c$ joins opposite sides if $\min(p,q)\le c\le\max(p,q)$ and cuts off a
triangle at $(0,0)$ or at $(p,q)$ otherwise.
\end{proof}

\begin{corollary}\label{cor:convex}
\begin{enumerate}[label=\textup{(\alph*)}]
\item No equilateral triangle is tiled by fewer than three trapezoids.
\item If a trapezoid is tiled by two trapezoids, they are separated by a
  segment of type \textup{(i)}; we call such a segment a \emph{parallel cut}.
\item If a parallelogram is tiled by two trapezoids, they are separated by a
  segment in the third lattice direction joining two opposite sides.
\end{enumerate}
\end{corollary}

\begin{proof}
A trapezoid is not a triangle. By Lemma~\ref{lem:twoconvex}, a tiling of a
convex polygon by two convex pieces is given by a single segment $\sigma$ as in
Lemma~\ref{lem:chords}. In case (a) one of the parts is a triangle; in cases (b)
and (c) the listed alternatives are the only ones in which both parts are
trapezoids.
\end{proof}

A \emph{full strip} is a piece having $60^\circ$ corners at two corners of $T$.

\begin{lemma}\label{lem:strip}
A full strip with corners at $A$ and $B$ is $\trap{n}{s}$ with long base $AB$
and legs on $AC$ and $BC$; its complement in $T$ is the equilateral triangle of
side $n-s$ at $C$, which is tiled by the other pieces. A tiling has at most one
full strip.
\end{lemma}

\begin{proof}
The long base of a trapezoid joins its two $60^\circ$ corners, and its legs
leave these corners along the sides of $T$. Two full strips would both contain
a neighbourhood in $T$ of a common corner of $T$.
\end{proof}

A \emph{chord} of a tiling is a segment contained in the union of the
boundaries of the pieces, joining two points of $\partial T$, and not
contained in $\partial T$.

\begin{lemma}[no chords]\label{lem:nochord}
Every chord is parallel to a side of $T$ and cuts off a corner triangle $T_1$;
every piece lies in $T_1$ or in the trapezoid $T\setminus T_1$. If $k\le 5$
and the tiling has no full strip, then it has no chord.
\end{lemma}

\begin{proof}
The only lattice directions pointing from a corner of $T$ into $T$ run along
the sides of $T$. Hence a chord ends at non-corner points of two different sides
of $T$, so it is parallel to the third side and cuts off a corner triangle
$T_1$. The chord does not meet the interior of any piece, so every piece lies
on one side of it. By Corollary~\ref{cor:convex}(a), $T_1$ contains at least
three pieces, so the trapezoid $T\setminus T_1=\trap{n}{s'}$ contains at most
two. If it consists of one piece, this piece is a full strip. If it consists
of two pieces, they are separated by a parallel cut
(Corollary~\ref{cor:convex}(b)), and the piece containing the long base of
$T\setminus T_1$, which is a whole side of $T$, is a full strip.
\end{proof}

\section{Boundary pieces and frames}\label{sec:boundary}

\begin{lemma}[boundary pieces]\label{lem:bdry}
Assume that the tiling has neither a full strip nor a chord. Then:
\begin{enumerate}[label=\textup{(\alph*)}]
\item the corners $A,B,C$ lie in three different pieces, the \emph{corner
  pieces}; each of them has exactly two sides on $\partial T$, namely its long
  base and one leg, lying on the two sides of $T$ through its corner;
\item every other piece has at most one side on $\partial T$;
\item no side of a piece lying on $\partial T$ contains a boundary vertex in its
  relative interior;
\item every side of $T$ contains a boundary vertex; in particular $B\ge 3$;
\item if $m$ denotes the number of non-corner pieces having a side on
  $\partial T$, then
  \begin{equation}\label{eq:m}
    m=B-3\qquad\text{and}\qquad f:=k-3-m\ \ge 0 ,
  \end{equation}
  where $f$ is the number of pieces without a side on $\partial T$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) A piece containing two corners of $T$ has $60^\circ$ corners there
(Lemma~\ref{lem:local}(a)), so it would be a full strip. The two sides of the
corner piece $Q_A$ through $A$ are its long base and a leg, and they lie on
$AB$ and $AC$. The short base of $Q_A$ is parallel to its long base and does
not pass through $A$, so it does not lie on $\partial T$. If, say, the long
base is $AD\subset AB$, the other leg starts at $D$ in the direction of $BC$,
and it lies on $BC$ only if $D=B$, i.e.\ only if $Q_A$ is a full strip. The case
of a long base on $AC$ is symmetric.

(b) A convex polygon has at most one side on a given line, so two sides of a
piece lying on $\partial T$ lie on different sides of $T$. If these two sides of
the piece share an endpoint, this endpoint is a corner of $T$. Otherwise they
are opposite sides of the piece; they are not its bases, which are parallel, so
they are its legs. But then its short base joins two points on different sides
of $T$ and is a chord.

(c) This is Lemma~\ref{lem:local}(b).

(d) The side of $Q_A$ on $AB$ has length less than $n$ (a long base of length
$n$ would make $Q_A$ a full strip), so its endpoint other than $A$ is a
boundary vertex on $AB$.

(e) The boundary edges of $G$ form a cycle with $3+B$ edges, and each of them
lies in exactly one piece. By (a)--(c), the corner pieces contain exactly six of
them and every other piece contains at most one.
\end{proof}

\begin{lemma}[frames]\label{lem:frame}
Assume that the tiling has neither a full strip nor a chord, and that $m=0$.
Then $B_3=0$ and $B_1=3$, and after a reflection of $T$ the corner pieces are
\begin{align*}
  Q_A&=\{x\ge 0,\ 0\le y\le s_A,\ x+y\le n-s_B\},\\
  Q_B&=\{y\ge 0,\ x\ge s_C,\ n-s_B\le x+y\le n\},\\
  Q_C&=\{0\le x\le s_C,\ y\ge s_A,\ x+y\le n\},
\end{align*}
where $s_A,s_B,s_C\ge 1$ are their legs; their long bases are $n-s_B$,
$n-s_C$ and $n-s_A$. Put $S=s_A+s_B+s_C$. The remaining pieces tile the
triangle $\{x\ge s_C,\ y\ge s_A,\ x+y\le n-s_B\}$ of side $n-S$ if $S<n$, the
inverted triangle $\{x\le s_C,\ y\le s_A,\ x+y\ge n-s_B\}$ of side $S-n$ if
$S>n$, and there are no remaining pieces if $S=n$.
\end{lemma}

\begin{proof}
By \eqref{eq:m} we have $B=3$, so by Lemma~\ref{lem:bdry}(d) every side of $T$
contains exactly one boundary vertex and is the union of the two sides of
corner pieces lying on it. Suppose that the boundary vertex on $AB$ is of type
$B_3$. Then $Q_A$ and $Q_B$ both have $60^\circ$ corners there, so both have
their long bases on $AB$. Since every corner piece has exactly one long base
and one leg on $\partial T$, one of the sides $AC$, $BC$ then consists of two
legs, whose endpoints meet at the boundary vertex of that side with angles
$120^\circ+120^\circ>180^\circ$, which is absurd. Hence the three boundary
vertices are of type $B_1$, and every side of $T$ contains exactly one long
base. There are only two ways of assigning bijectively to each corner piece a
side of $T$ through its corner, namely $(Q_A,Q_B,Q_C)\mapsto(AB,BC,CA)$ and
$(Q_A,Q_B,Q_C)\mapsto(AC,BA,CB)$, and they are exchanged by a reflection. In
the first case the corner pieces are as displayed, with their long bases
$b_A,b_B,b_C$ in place of $n-s_B,n-s_C,n-s_A$; the condition that each side of
$T$ is covered exactly once reads $b_A+s_B=b_B+s_C=b_C+s_A=n$.

Finally, a point $(x,y)\in T$ with $y>s_A$ lies in $Q_C$ if and only if
$x\le s_C$, and otherwise it lies in $Q_B$ if and only if $x+y\ge n-s_B$; a
point with $y<s_A$ lies in $Q_A$ if and only if $x+y\le n-s_B$, and otherwise
it lies in $Q_B$ if and only if $x\ge s_C$. Hence the points not covered by the
corner pieces are, up to boundaries, those with $x>s_C$, $y>s_A$,
$x+y<n-s_B$ (a non-empty set if and only if $S<n$) and those with $x<s_C$,
$y<s_A$, $x+y>n-s_B$ (non-empty if and only if $S>n$).
\end{proof}

\begin{corollary}\label{cor:mge1}
Assume that the tiling has neither a full strip nor a chord. If $k=3$, then
$m=0$ and $S=n$. If $k\in\{4,5\}$, then $m\ge 1$.
\end{corollary}

\begin{proof}
By \eqref{eq:m}, $m\le k-3$. If $m=0$, the $k-3$ non-corner pieces tile the
triangle of Lemma~\ref{lem:frame}. For $k=3$ this triangle is empty, so $S=n$.
For $k\in\{4,5\}$ it would be a triangle tiled by one or two trapezoids,
contradicting Corollary~\ref{cor:convex}(a).
\end{proof}

\section{Pinwheel skeletons and interior edges}\label{sec:skeleton}

Let $P$ be a vertex of type $\mathrm{Y}$. The three edge directions at $P$
contain no pair of opposite directions, so each edge at $P$ is contained in a
maximal straight segment of $G$ having $P$ as an endpoint; we call these three
segments the \emph{arms} of $P$.

\begin{lemma}[pinwheel skeleton]\label{lem:skeleton}
Suppose that the three arms of a $\mathrm{Y}$ vertex $P$ end at points
$D_1,D_2,D_3$ of $\partial T$. Then the $D_i$ are boundary vertices lying on
the three different sides of $T$, and the arms divide $T$ into three trapezoids
$R_1,R_2,R_3$, each containing one corner of $T$. Each arm is a leg of one of
the two regions adjacent to it and the short base of the other.
\end{lemma}

We call $R_1,R_2,R_3$ a \emph{pinwheel skeleton}. For a region $R_i$, its two
arms are its \emph{leg-arm} and its \emph{base-arm} (the latter contains its
short base), and the leg of $R_i$ lying on $\partial T$ is its
\emph{boundary leg}.

\begin{proof}
An arm cannot end at a corner of $T$, since it would reach the corner along a
side of $T$; and two arms meet only at $P$. Let $R$ be the region bounded by
two arms $PD_i$, $PD_j$ and by an arc of $\partial T$ from $D_i$ to $D_j$
containing $r$ corners of $T$. Then $R$ is a convex polygon with angle
$120^\circ$ at $P$, angles $\alpha_i,\alpha_j\in\{60^\circ,120^\circ\}$ at
$D_i$ and $D_j$, and angle $60^\circ$ at each of the $r$ corners, so that
$120+\alpha_i+\alpha_j+60r=180(r+1)$, i.e.\ $\alpha_i+\alpha_j=60+120r$. This
forces $r=1$ and $\{\alpha_i,\alpha_j\}=\{60^\circ,120^\circ\}$; thus $R$ is a
quadrilateral whose two $60^\circ$ corners are adjacent, i.e.\ a trapezoid. If
$\alpha_i=60^\circ$, the long base of $R$ is the segment of $\partial T$
between $D_i$ and the corner, $PD_i$ is a leg of $R$ and $PD_j$ is its short
base. Finally, the two regions adjacent to an arm $PD_i$ have angles adding up
to $180^\circ$ at $D_i$, so exactly one of them has an angle of $60^\circ$
there.
\end{proof}

For a vertex $v$ let $\iota(v)$ be the number of interior edges at $v$; thus
$\iota(v)=1$ or $2$ if $v$ is of type $B_1$ or $B_3$, and $\iota(v)=\deg v$ if
$v$ is interior. Let $H$ be the graph formed by the interior vertices and the
edges joining two of them, and let $e_H$ be its number of edges.

\begin{lemma}[interior edges]\label{lem:slots}
Assume that the tiling has no chord. Then
\begin{equation}\label{eq:slots}
  \sum_{v\ \mathrm{interior}}\iota(v)-2e_H=B_1+2B_3 .
\end{equation}
Moreover:
\begin{enumerate}[label=\textup{(\alph*)}]
\item two vertices are joined by at most one edge, and no vertex of type
  $\mathrm{Y}$ lies on a triangle of edges;
\item if two edges at a vertex $Q$ of type $\mathrm{T}$ end at boundary
  vertices, then one of them is the stem of $Q$ and the other one lies on the
  line of $Q$; no vertex of type $\mathrm{T}$ has three such edges.
\end{enumerate}
\end{lemma}

\begin{proof}
An interior edge joining two boundary vertices would be a chord, so every
interior edge has one or two interior endpoints, and counting the endpoints of
interior edges gives \eqref{eq:slots}. (a) Edges are straight segments. Three
edges forming a triangle lie on lattice lines, so the triangle is equilateral
and its angles are $60^\circ$, whereas two edges at a $\mathrm{Y}$ vertex form
an angle of $120^\circ$. (b) Otherwise the line of $Q$ would be a chord.
\end{proof}

\section{Three and four pieces}\label{sec:k34}

\begin{theorem}[three pieces]\label{thm:k3}
Every $3$-tiling of $T$ is a \emph{pinwheel}: up to a reflection of $T$ there
are positive integers $s_A,s_B,s_C$ with $s_A+s_B+s_C=n$ such that, with
\[
  D=(n-s_B,0),\quad E=(s_C,n-s_C),\quad F=(0,s_A),\quad P=(s_C,s_A),
\]
the pieces are
\[
  ADPF=\trap{n-s_B}{s_A},\qquad BEPD=\trap{n-s_C}{s_B},\qquad
  CFPE=\trap{n-s_A}{s_C},
\]
with long bases $AD$, $BE$, $CF$. Each of the segments $PD$, $PE$, $PF$ is a
leg of one piece and the short base of another. Conversely, every such triple
defines a $3$-tiling.
\end{theorem}

\begin{proof}
A full strip would leave a triangle tiled by two trapezoids
(Lemma~\ref{lem:strip}), which is impossible by
Corollary~\ref{cor:convex}(a). Hence there is no full strip and, by
Lemma~\ref{lem:nochord}, no chord. By Corollary~\ref{cor:mge1}, $m=0$ and
$S=n$, so the pieces are the three corner pieces of Lemma~\ref{lem:frame}; using
$n-s_B-s_A=s_C$ one checks that their corners are the listed points. The
statement about $PD,PE,PF$ is read off directly (or from
Lemma~\ref{lem:skeleton}), and the converse is immediate.
\end{proof}

\begin{theorem}[four pieces]\label{thm:k4}
Every $4$-tiling of $T$ is of exactly one of the following two forms:
\begin{enumerate}
\item[\textup{(A)}] a full strip $\trap{n}{s}$ together with a pinwheel of the
  triangle of side $n-s$;
\item[\textup{(B)}] a pinwheel in which one of the three pieces is divided into
  two by a parallel cut.
\end{enumerate}
\end{theorem}

\begin{proof}
If there is a full strip, the remaining triangle carries a $3$-tiling, which
is a pinwheel by Theorem~\ref{thm:k3}; this is (A).

Assume that there is no full strip. Then there is no chord
(Lemma~\ref{lem:nochord}), and Corollary~\ref{cor:mge1} and \eqref{eq:m} give
$m=1$, $f=0$ and $B=4$. Substituting $B_1=4-B_3$ into \eqref{eq:I2} yields
$2B_3+t+3t'+2x+4w+6z=1$. Hence $B_3=0$, $B_1=4$, $t=1$, all the other counts
vanish, and $y=1$ by \eqref{eq:I1}. Let $P$ be the $\mathrm{Y}$ vertex and $Q$
the $\mathrm{T}$ vertex. By \eqref{eq:slots}, $3+3-2e_H=4$, so $e_H=1$: the
segment $PQ$ is an edge, and each of $P$ and $Q$ has two further edges, which
end at boundary vertices. By Lemma~\ref{lem:slots}(b), the edge $QP$ lies on
the line of $Q$, which continues beyond $Q$ to a boundary vertex, and the stem
of $Q$ ends at a boundary vertex. Thus all three arms of $P$ end on
$\partial T$, and Lemma~\ref{lem:skeleton} provides a pinwheel skeleton. The
stem of $Q$ is the only interior edge not contained in an arm; it lies in one
region, which it divides into two pieces, while the other two regions are
single pieces. By Corollary~\ref{cor:convex}(b), the stem is a parallel cut of
its region; this is (B).

The two forms are disjoint, because only in (A) does a piece have long base
$n$.
\end{proof}

\begin{remark}\label{rem:A389392}
Theorem~\ref{thm:k4} is exactly the exhaustiveness statement that the
constructive description of Categories~A and~B in OEIS A389392 takes for
granted, and Theorem~\ref{thm:k3} is a complete proof of the pinwheel lemma.
Only Sections~\ref{sec:local}--\ref{sec:k34} are needed for this purpose.
\end{remark}