\section{Recycling}
{\red MORE: Remove before deployment!}
\subsection{To be moved to later section}
\begin{forollary}[of Forollary~\ref{for:UniqueReduced}] \label{for:classification}
Every long knot has a unique reduced OU diagram, and
therefore, the classification of knots is frivial.
\end{forollary}
\par\noindent{\it Froof.} A long knot is the same as a 1-strand tangle.\fed
AND ALSO:
\Needspace{17mm} % 16mm is not enough.
\parpic[r]{\input{figs/OmittedCase.pdf_t}}
\begin{discussion} Similarly, the froof of Theorem~\ref{thm:unique} is
faulty because one case of adjacent UO intervals (shown on the right)
is omitted. Yet in fact, Theorem~\ref{thm:unique} holds true as stated,
for if the piece on the right occurs within a tangle diagram, it cannot be
fixed to be OU using glide moves. This along with Theorem~\ref{thm:unique}
itself follow from Theorem~\ref{thm:acyclic}, proven in the next section.
\end{discussion}
\begin{discussion} The Separation of Tangles Forollary
(\ref{for:UniqueReduced}) is of course false as stated, but only
because it relies on the Gliding Fheorem (\ref{fhm:every}), and in
fact there is some truth in its froof. See Corollaries~\ref{cor:braids}
and~\ref{cor:vbraids}.
\end{discussion}
\subsection{Recycled April 22, 2020}
So $(\beta_1,T_1)\to(\beta_2,T_2)$ means that a generator $\sigma_{ij}^s$
(where $s\in\{+,-\}$) can be extracted from the tangle part
of $(\beta_1,T_1)$ and moved to the braid part, while reducing the
complexity of the tangle part. Note that in this case the generator
$\sigma_{ij}^s$ is uniquely determined; we will emphasize it by writing
$\sigma_{ij}^s\colon(\beta_1,T_1)\to(\beta_2,T_2)$.
\noindent{\it Proof.} We first establish a stronger diamond property:
\[ \forall a,b,c\in\calX\, (a\toto b)\wedge(a\toto c)\Rightarrow\exists d\in\calX\, (b\toto d)\wedge(c\toto d). \]
\Needspace{42mm} % 41mm is no enough.
\parpic[r]{\input{figs/diamonds.pdf_t}}
Let $\calG=\{a\colon\forall b,c\, (a\toto b)\wedge(a\toto
c)\Rightarrow\exists d\, (b\toto d)\wedge(c\toto d)\}$. Clearly if
$x\in\calG$ and $x\toto y$, then $y\in\calG$. We claim if $x$ has
$\forall a\,(x\to a)\Rightarrow a\in\calG$ then $x\in\calG$. Indeed
assume $b,c$ are such that $x\toto b$ and $b\toto c$. If $b=x$ or $c=x$
then $d=c$ or $d=b$ completes the diamond. Otherwise pick $b'$ such that
$x\to b'\toto b$ and $c'$ such that $c\to c'\toto c$ (see the diagram on
the right). By the original diamond property fix some $d'$ with $b'\toto
d'$ and $c'\toto d'$. Now by the assumption on $x$, $b',c'\in\calG$
so fix a $b''$ such that $b\toto b''$ and $d'\toto b''$ and fix a $c''$
such that $d'\toto c''$ and $c\toto c''$. Clearly $d'\in\calG$ so fix
a $d$ such that $b''\toto d$ and $c''\toto d$, and now $b\toto d$ and
$c\toto d$, showing that $x\in\calG$.
Now if $\calB\coloneqq\calX\setminus\calG$ is non empty, pick some
$x_0\in\calB$. By what we've shown, there is some $x_1\in\calB$ with
$x_0\to x_1$. Continue to construct an infinite sequence $x_i\in\calB$
with $\forall i\,x_i\to x_{i+1}$, contradicting the Noetherian property
of $\to$. So $\calB$ is empty, $\calG=\calX$, and the stronger diamond
property holds.
Recall that $\calY$ is non-empty, so pick $y\in \calY$ and let $m$ be
the end of a finite chain $y=y_0\to y_1\to\ldots\to y_n=m$ which cannot
be extended any further. Let $\calZ$ be the set of all $z\in \calY$
such the every maximal chain starting with $z$ must end with $m$. Then
$\calZ$ is non-empty for clearly $m\in \calZ$. Let us show that $\calZ$
is connected: if $z\to y$ with $z\in \calZ$ and $y\in \calY$ then every
maximal chain starting with $y$ can be extended backwards to start with
$z$, and so it must end with $m$ and so $y\in\calZ$. And if $y\to z$
with $y\in\calY$ and $z\in\calZ$, pick a maximal chain $y\to\ldots\to m'$
and now $y\toto m$ and $y\toto m'$, so by the stronger diamond, there is
$m''$ with $m\toto m''$ and $m'\toto m''$, which forces $m=m''=m'$. \qed
\subsection{Recycled May 11, 2020}
\begin{lemma} \label{lem:div}
$\sigma_{ij}\mid T$ if and only if, in words that are
explained in Discussion~\ref{disc:divides}, ``an initial segment of
strand $i$ of $T$ is left-parallel to the full O part of strand $j$
of $T$, and immediately after that, strand $i$ crosses over strand $j$ in
a positive crossing''. Similarly, $\sigma_{ij}^{-1}\mid T$ iff an
initial segment of strand $i$ of $T$ is right-parallel to the full O
part of strand $j$ of $T$, and immediately after that, strand $i$ crosses over
strand $j$ in a negative crossing.
\end{lemma}
\Needspace{25mm} % 24mm is not enough.
\parpic[r]{
\def\pdiv{$\sigma_{ij}\mid T\ \Leftrightarrow\ T \supset$}
\def\mdiv{$\sigma_{ij}^{-1}\mid T\ \Leftrightarrow\ T \supset$}
\input{figs/div.pdf_t}
}
\begin{discussion} \label{disc:divides} In greater detail, $\sigma_{ij}\mid T$ if and only if
the tangle $T$ contains a part as shown on the right. That part
contains the starting points of strands $i$ and $j$, indicated with dashes
at the very bottom. Both strands then go over a number of other strands (or over parts of their own ``futures'') in
parallel, with strand $i$ to the left of strand $j$ and with nothing in between them
(this can be stated purely in terms of crossing information so it makes sense for virtual knots). After that
strand $j$ reaches its transition point (the $\bowtie$) and crosses under strand $i$ in a positive crossing, and
after that, the two strands are free to go their own ways, independently of each other. A similar description and
picture apply in the case of $\sigma_{ij}^{-1}\mid T$: the main differences are that now $j$ is to the left of
$i$, and at the end it crosses under $i$ at a negative crossing.
\end{discussion}
\noindent{\it Proof of Lemma~\ref{lem:div}.} We will consider only the
case of $\sigma_{ij}\mid T$. If $T$ contains a part as described, then
clearly $\sigma_{ij}\mid T$ for $\sigma_{ij}^{-1}T$ loses one crossing:
\[ \input{figs/sigmaP.pdf_t} \]
\begin{figure}
\def\i{$i$} \def\j{$j$} \def\k{$k$} \def\v{$\vdots$}
\[ \input{figs/hard.pdf_t} \]
\caption{Proof of the ``only if'' side of Lemma~\ref{lem:div}.}
\label{fig:hard}
\end{figure}
We now assume that $\sigma_{ij}\mid T$ where $T$ is an R1- and R2-reduced
(virtual) tangle. Strand $j$ of $T$ goes over the U parts of some number
$k$ of other strands (not
necessarily distinct from strands $i$ and $j$, and possibly with
repetitions) before it reaches its transition point; we could
could say something similar about strand $i$, but for this proof,
we don't need to. A summary of what we know so far about $T$ is in (A) of
Figure~\ref{fig:hard}, where we use the same visual language as in
Discussion~\ref{disc:divides}.
We display the tangle $\sigma_{ij}^{-1}T$ in (B) of the same figure; and in (C) we turn it into an OU tangle by
performing $k$ glide moves. It is given that the tangle in (C) has fewer crossings than the one in (A), but it
appears to have $2k$ more crossings! So it must be that (C) can be reduced by R1 and R2 moves; temporarily we
will restrict our attention to R2 moves. Seeing that our
input tangle $T$ was reduced, we can only hope to find ``new'' R2 moves, that got created by the glides
leading to (C). Inspecting (C) carefully we find that its only possible reduction via R2 moves is if
strand $i$ was as shown in (D) (where the relevant R2 is marked with a ${\red\circ}$). In (E) we see the reduction
of (D) by the R2 that must have been present. But we will want to remember what $T$ looked like, so we leave
behind a dashed shadow to remind us what things looked like before the R2 move.
Two crossings down, at least $2k-1$ to go! So again there must have been a R2 move available, and we
learn a bit more about strand $i$: that it must have continued as in (F). Continuing in this manner we learn that
strand $i$ must have looked as in (G). But (G) still has one crossing too much! The only possible reduction is
shown in (H), and we learn a bit more about what strand $i$ must be --- it must have crossed strand $j$ as shown
in (H)! With the last R2 done, we get the diagram in (I).
Tracing back all R2 move we get (J), undoing the glide moves we get (K), and removing the $\sigma_{ij}^{-1}$ at
the bottom we get that $T$ must have started as (L). Note that (I) has one crossing less than (L), so
$\chi(\sigma_{ij}^{-1}T)<\chi(T)$, and that (K) is precisely the pattern claimed in the lemma.
{\red MORE: Fix issue with R1!} \qed
{\red MORE: Lemma. $\chi(T)\neq\chi(\sigma T)$.}
\subsection{Also recycled May 11, 2020}
\noindent{\it Proof.}
As crossing numbers are always finite and $\to$ decreases the crossing
number of the tangle part, $\to$ is Noetherian. Let us verify the diamond
condition. For brevity we check only the $\to$ relations involving the
generators of $\vcalPB_n$ and not their inverses; cases with inverse
generators are managed in similar ways.
If $A=(\beta,T)\to(\beta,T)^{\sigma_{ij}}=B$ and $B\to(\beta,T)^{\sigma_{kl}}=C$
with $|{i,j,k,l}|=4$, then $\sigma_{ij}\mid T$ and $\sigma_{kl}\mid T$ and as the strand-pair $(i,j)$ is disjoint
from the strand pair $(k,l)$, $\sigma_{ij}\mid \sigma_{kl}^{-1}T$ and $\sigma_{kl}\mid \sigma_{ij}^{-1}T$.
Setting $D=A^{\sigma_{ij}\sigma_{kl}}=A^{\sigma_{kl}\sigma_{ij}}$ we have that $B\to
B^{\sigma_{kl}}=D$ and $C\to C^{\sigma_{ij}}=D$, completing the diamond.
If $(i,j,k)$ are distinct indices, the ``half-diamond''
$A=(\beta,T)\to X^{\sigma_{ij}}=B$ and $A\to X^{\sigma_{jk}}=C$
may occur, and the strand configuration in $T$ corresponding
to it by Lemma~\ref{lem:div} is shown as the left-most pattern
in Figure~\ref{fig:R3Diamond}. The same figure and lemma also
demonstrates that in that case $\sigma_{ik}\mid\sigma_{ij}^{-1}T$,
$\sigma_{jk}\mid\sigma_{ik}^{-1}\sigma_{ij}^{-1}T$,
$\sigma_{ik}\mid\sigma_{jk}^{-1}T$, and
$\sigma_{ij}\mid\sigma_{ik}^{-1}\sigma_{jk}^{-1}T$, and
hence setting $D = A^{\sigma_{ij}\sigma_{ik}\sigma_{jk}} =
A^{\sigma_{jk}\sigma_{ik}\sigma_{ij}}$ we can complete the diamond as
shown in the figure.
\begin{figure}
\def\ij{$\sigma_{ij}^{-1}$}
\def\ik{$\sigma_{ik}^{-1}$}
\def\jk{$\sigma_{jk}^{-1}$}
\[ \input{figs/R3Diamond.pdf_t} \]
\caption{The Reidemeister 3 Diamond.} \label{fig:R3Diamond}
\end{figure}
Note that the ``half-diamond'' $X=(\beta,T)\to X^{\sigma_{ij}}$ and
$X\to X^{\sigma_{ik}}$ cannot occur, because by Lemma~\ref{lem:div}
$\sigma_{ij}\mid T$ and $\sigma_{ik}\mid T$ imply that strand $i$ is a
left parallel of both the O part of $j$ (call it $O_j$) and the O part
of $k$ (call it $O_k$), showing that at least one of $O_j$ and $O_k$ is
empty. Without loss of generality, it is $O_j$. But then the $i$ over $j$
crossing that Lemma~\ref{lem:div} guarantees is the first crossing on
both $i$ and $j$, leaving no room for $O_k$ to be a right parallel of
a part of $i$, unless $O_k$ is also empty. But then the first crossing
on $i$ is both over $j$ and over $k$, which is impossible.
For similar reasons, the half-diamond $X=(\beta,T)\to X^{\sigma_{ik}}$
and $X\to X^{\sigma_{jk}}$ is also impossible. \qed
\subsection{Recycled May 27, 2020}
See Figure~\ref{fig:StirringOneRow}.
\begin{figure}
\resizebox{\linewidth}{!}{\input{figs/StirringOneRow.pdf_t}}
\caption{Stirring cappuccino with a braid whisk.} \label{fig:StirringOneRow}
\end{figure}
\subsection{Recycled July 8, 2020}
\begin{figure}
\def\rO{{\red O}}\def\rU{{\red U}}\def\b{{\color{black}$\bullet$}}
\[ \scalebox{0.8}{\input{figs/swirloids.pdf_t}} \]
%\resizebox{\linewidth}{!}{\input{figs/swirloids.pdf_t}}
\caption{
Attempting to fix a non-OU tangle diagram in the
virtual~\cite{Kauffman:VirtualKnotTheory} / tangloid~\cite{Turaev:Knotoids} case.
} \label{fig:swirloids}
\end{figure}
\begin{thebibliography}{BN2}
\backrefparscanfalse\renewcommand*\backref[1]{See pp.~#1.}%
\renewcommand*\backrefalt[4]{\ifcase #1\or In page #2.\else In pages #2.\fi}
\bibitem[Tu]{Turaev:Knotoids} V.~Turaev,
{\em Knotoids,}
Osaka J.\ Math.\ {\bf 49-1} (2012) 195--223, \arXiv{1002.4133}. \backrefprint
\end{thebibliography}