\section{The hulls of diagram 5}\label{app:d5hulls} We prove Theorem~\ref{thm:hulls}, which gives the hulls of \S\ref{ss:d5} explicitly. Throughout, $P$ is a place of diagram~$5$ with $zw$-triangle $T$ and outer bodies $A$, $B$. The points $z_c$, $w_c$ are the centres of mass of $T$. We use the coordinates $y=(x_{At_0},x_{At},x_{Bt_0},x_{Bt},x_{AB})$ of \S\ref{ss:d5}. We put $e_A=(1,1,0,0,0)$ and $e_B=(0,0,1,1,0)$. \begin{theorem}\label{thm:hulls} Consider the model of $\Vtwo^{\mathrm w}$, $\Vtwo$ or $\VN$ \textup{(}\S\ref{ss:d5}\textup{)} for a pair of modes. The projection to $y$ of each point of the model lies in the polytope that Table~\ref{tab:d5hulls} gives for this hull and this pair of modes, up to the exchanges of Lemma~\ref{lem:d5sym}; where the table gives $\emptyset$, the model is empty. In the centred case, the model of Theorem~C0 of the supplementary material is empty for the pairs of modes \textup{(Z1,\,Zs$_{t_0}$)} and \textup{(W1,\,Ws$_{t_0}$)} and their exchanges, where $t_0$ is the body of $T$ at the centre of mass of $T$. \end{theorem} \begin{table}[ht] \centering \small \begin{tabular}{@{}llll@{}} \toprule pair of modes & $\Vtwo^{\mathrm w}$ & $\Vtwo$ & $\VN$\\ \midrule (F,\,F) & $\conv\{0,e_A,e_B\}$ & $\{0\}$ & $\{0\}$\\ (F,\,Z1) & $\conv\{0,e_A,e_B\}$ & $[0,e_B]$ & $\emptyset$\\ (Z1,\,Z1) & $P_{11}$ & $P_{11}$ & $[-\frac12(e_A+e_B),0]$\\ (Z1,\,W1) & $[e_A,e_B]$ & $[e_A,e_B]$ & $\emptyset$\\ (Z1,\,Zs) & $P_{1s}$ & $P_{1s}$ & $\emptyset$\\ (Zs,\,Zs) & $P_{ss}$ & $P_{ss}$ & $P_{ss}$\\ (F,\,Zs), (Z1,\,Ws), (Zs,\,Ws) & $\emptyset$ & $\emptyset$ & $\emptyset$\\ \bottomrule \end{tabular} \caption{The hulls of \S\ref{ss:d5} for nine pairs of modes, in the coordinates $y$. Here $P_{11}=\{ae_A+be_B:\ -\frac12\le a,b\le1,\ a+b\le1\}$, with the vertices $-\frac12(e_A+e_B)$, $-\frac12e_A+e_B$, $e_B$, $e_A-\frac12e_B$, $e_A$. The polytope $P_{1s}$, of dimension $4$, has the vertices $-\frac12(e_A+e_B)$, $(-\frac12,-\frac12,0,-\frac12,0)$, $0$, $e_A-\frac12e_B$, $(1,1,0,-\frac12,-\frac12)$, $(1,1,0,-\frac12,0)$, $e_A$, $(1,1,\frac13,0,\frac13)$. The polytope $P_{ss}$, of dimension $5$, has the vertices $-\frac12(e_A+e_B)$, $(-\frac12,-\frac12,0,-\frac12,0)$, $(-\frac13,-\frac13,0,-\frac12,0)$, $(0,-\frac12,-\frac12,-\frac12,0)$, $(0,-\frac12,-\frac13,-\frac13,0)$, $(0,-\frac12,0,-\frac12,0)$, $0$, $(\frac13,0,\frac13,0,\frac13)$.}\label{tab:d5hulls} \end{table} \subsection{Orders and relations}\label{ss:d5rel} For $X\in\{A,B\}$ let $Y$ be the other outer body. We write $c_X=e(z_X-z_c)$, $d_X=e(w_X-w_c)$, $p^z_X=e(z_X)$, $p^w_X=e(w_X)$, $p^z_c=e(z_c)$, $p^w_c=e(w_c)$ and \[ t^z_X=e\Big(\sum_{t\in T}m_tZ_{tX}\Big),\qquad t^w_X=e\Big(\sum_{t\in T}m_tW_{tX}\Big), \] and we put $\zeta=\zeta_{AB}$, $\omega=\omega_{AB}$ and $r=x_{AB}$, so that $e(Z_{AB})=\zeta-3r$ and $e(W_{AB})=\omega-3r$. A point of a model is a real vector, with one entry for each modelled order, that satisfies the relations of the model. We denote its entries by the names of the orders. Inside $T$ the differences $z_t-z_c$ and $w_t-w_c$ have exponent at least $1$. So, by Lemma~\ref{lem:nearcluster}, a body $X$ in mode F has $c_X,d_X<1$ and $x_{Xt}=\bar x_X:=\frac12(c_X+d_X)$ for all $t\in T$. A body $X$ in mode Z1 or Zs$_{t_0}$ has $c_X\ge1$, the far order $f_X:=d_X<1$ and the order $s_X=\zeta_{Xt_0}$, which is $1$ in mode Z1. In mode Zs$_{t_0}$, $s_X>1$, $s_X+3f_X<4$ and $3s_X+f_X<4$. Then $x_{Xt}=\bar x_X:=\frac12(1+f_X)$ for $t\ne t_0$, and $x_{Xt_0}=x^0_X:=\frac12(s_X+f_X)$. In the modes F and Z1 we put $x^0_X=\bar x_X$, so that $y=(x^0_A,\bar x_A,x^0_B,\bar x_B,r)$. The modes W1 and Ws$_{t_0}$ are the same with the colours exchanged. The orders $c_X$, $d_X$, $\zeta$, $\omega$ and those of the positions are at least $-2$, since all components of $\cZ$ and $\cW$ have exponent at least $-2$ (\S\ref{ss:orders}). So $\bar x_X\ge-\frac12$ in the modes other than F. As $A$ and $B$ have no strokes, $\zeta+3\omega<4$ and $3\zeta+\omega<4$. The terms of $\sum_tm_tZ_{tX}$ and $\sum_tm_tW_{tX}$ have the orders $\zeta_{Xt}-3x_{Xt}$ and $\omega_{Xt}-3x_{Xt}$. In mode Zs$_{t_0}$ the term of $t_0$ has the least order. These two facts give the relations (S): \begin{enumerate}[leftmargin=3.4em,labelwidth=2.8em,labelsep=0.6em,align=left] \item[(S)] In mode F, $t^z_X\ge c_X-3\bar x_X$ and $t^w_X\ge d_X-3\bar x_X$, with equality in the models of $\Vtwo$ and $\VN$. In mode Z1, $t^z_X\ge1-3\bar x_X$ and $t^w_X\ge f_X-3\bar x_X$. In mode Zs$_{t_0}$, $t^z_X=s_X-3x^0_X$ and $t^w_X=f_X-3x^0_X$. \end{enumerate} Each of the following relations says that, among the orders of the terms of an exact identity with nonzero constant coefficients, the minimum is attained at least twice. We list the orders and then the identity. \begin{enumerate}[leftmargin=3.4em,labelwidth=2.8em,labelsep=0.6em,align=left] \item[(P)] Any three of the six orders $c_A$, $c_B$, $\zeta$, $p^z_A$, $p^z_B$, $p^z_c$, and likewise in $w$. By the centre of mass, $Mz_c+m_A(z_A-z_c)+m_B(z_B-z_c)=0$. So the six quantities $z_A-z_c$, $z_B-z_c$, $z_{AB}$, $z_A$, $z_B$, $z_c$ are linear combinations of the first two and no two of them are proportional. \item[(E)] $p^z_X$, $\zeta-3r$ and $t^z_X$, and likewise in $w$: the equation $z_X=m_YZ_{XY}-\sum_tm_tZ_{tX}$ of \eqref{eq:AK4}. \item[(F)] For $X$ not in mode F: $c_X$, $\zeta+\beta$, $t^z_X$, and also $d_X$, $\omega+\beta$, $t^w_X$, where $\beta$ is the order of $1-Mp_{AB}^3$. The identity is $M_T(z_X-z_c)=-M\sum_tm_tZ_{tX}-m_Y(1-Mp_{AB}^3)\,z_{XY}$, with its counterpart in $w$. \item[(I)] $2r$, $c_A+d_A$, $c_B+d_B$ and $0$: the identity $MI=MI_T+M_T\sum_Xm_X(z_X-z_c)(w_X-w_c)+m_Am_Bz_{AB}w_{AB}$. Here $I_T=\sum_tm_t(z_t-z_c)(w_t-w_c)$ has order at least $2$. The quantity $I$ is a nonzero constant on $C$ (Lemma~\ref{lem:U}). \item[(V)] $d_A+t^z_A$, $d_B+t^z_B$, $-r$ and $0$: the identity $U=I_T-\sum_Xm_X(w_X-w_c)\sum_tm_tZ_{tX}+m_Am_Bp_{AB}$. It follows from $p_{kl}=Z_{kl}w_{kl}$ and from the equations of the bodies $t\in T$ multiplied by $m_t(w_t-w_c)$. Here $U=I$ is a nonzero constant. \item[(V$'$)] $c_A+t^w_A$, $c_B+t^w_B$, $-r$ and $0$: the same identity with the colours exchanged. \end{enumerate} In (I), (V) and (V$'$) we leave out the term $I_T$, since the minimum is at most $0<2$. We also use the following relations, in which $X$ is in mode Z1. \begin{enumerate}[leftmargin=3.4em,labelwidth=2.8em,labelsep=0.6em,align=left] \item[(Q)] The torque of an outer body $X$ is $\Theta_X=\sum_{t\in T}m_t\big((z_t-z_c)W_{tX}-(w_t-w_c)Z_{tX}\big)$. Its order is $\tau_X$. We sum over $t\in T$ the equation of $w_t$ in \eqref{eq:AK4} multiplied by $m_t(z_t-z_c)$, minus that of $z_t$ multiplied by $m_t(w_t-w_c)$. The left sides give $w_c\sum_tm_t(z_t-z_c)-z_c\sum_tm_t(w_t-w_c)=0$ and the terms of the pairs inside $T$ cancel. Hence $m_A\Theta_A+m_B\Theta_B=0$ and $\tau_A=\tau_B$. For $X$ in mode Z1 the terms $m_t(z_t-z_c)W_{tX}$ have order at least $1+f_X-3\bar x_X=-\bar x_X$ and the terms $m_t(w_t-w_c)Z_{tX}$ order at least $2-3\bar x_X>-\bar x_X$. Thus $\tau_X\ge-\bar x_X$. \item[(R)] In the model of $\VN$: at most one of the following \emph{events} occurs: $c_X>1$; $t^z_X>1-3\bar x_X$; $t^w_X>f_X-3\bar x_X$; $\tau_X>-\bar x_X$; and, if $Y$ is in mode Zs, $\max(p^z_A,p^z_B,p^z_c)>1$ and $\tilde t_X>-\bar x_X$. This is the content of (R1) and (R2) of \S\ref{ss:d5}, which are proved in the supplementary material. \item[(T)] In the model of $\VN$, if $Y$ is in mode Zs$_{t_0}$: the orders $\tilde t_X$, $\tilde t_Y$ of the two quantities of (R3) (proved in the supplementary material) satisfy $\tilde t_X\ge-\bar x_X$, and $\tilde t_Y=\min(e_1,e_2)$ if $e_1\ne e_2$, where $e_1=1-\frac12(s_Y+3f_Y)$ and $e_2=-\bar x_Y$. The minimum of $\tilde t_X$, $\tilde t_Y$ and $p^w_c+1$ is attained at least twice. \end{enumerate} In the centred case, for $X$ in mode Z1 and $Y$ in mode Zs$_{t_0}$ with $t_0$ at the centre of mass of $T$, the model of the centred case in the supplementary material has the relations of the model of $\VN$ other than (T) and the event $\tilde t_X>-\bar x_X$, and moreover $c_Y>1$. In the modes W1 and Ws the relations are the same with the colours exchanged. The model of $\Vtwo$ is that of $\Vtwo^{\mathrm w}$ together with the equalities of (S) in mode F. The models of $\VN$ and of the centred case contain the relations of the model of $\Vtwo$. So a statement proved for the model of $\Vtwo^{\mathrm w}$ holds for the other models, and one proved for the model of $\Vtwo$ holds for those of $\VN$ and of the centred case. The models have further relations, which we do not use. \subsection{Symmetries and rules}\label{ss:d5rules} \begin{lemma}\label{lem:d5sym} The exchange of the colours maps the model of each hull, and that of the centred case, for a pair of modes onto the model of the same kind for the pair with Z1, Zs and W1, Ws exchanged, and it fixes $y$. The exchange of $A$ and $B$ maps the model for a pair of modes $(a,b)$ onto the model for $(b,a)$, and $y$ to $(y_3,y_4,y_1,y_2,y_5)$. Every pair of modes is the image of one of the nine pairs of Table~\ref{tab:d5hulls} under these exchanges. \end{lemma} \begin{proof} The system \eqref{eq:AK4} and the type are invariant under the exchange of $z$ and $w$. This exchange maps each relation of a model to a relation of the model for the pair of modes with the colours exchanged: (P), (E) and (F) in $z$ go to their counterparts in $w$, (V) goes to (V$'$), and (S), (Q), (R), (T) go to their mirror images. The exchange of $A$ and $B$ renames the outer bodies. The two exchanges generate a group of order $4$, whose orbits on the $25$ pairs of modes are those of the nine pairs of the table. \end{proof} By Lemma~\ref{lem:d5sym} we treat the nine pairs of Table~\ref{tab:d5hulls}. For their images under the exchange of the colours, the proofs below hold with (V$'$) in place of (V). \begin{lemma}\label{lem:d5rules} At every point of the model of $\Vtwo^{\mathrm w}$, for $X\in\{A,B\}$: \begin{enumerate}[label=\textup{(\alph*)}] \item In each colour, five of the six orders $c_A$, $c_B$, $\zeta$, $p^z_A$, $p^z_B$, $p^z_c$ are equal to their minimum $\mu_z$, and the sixth, the \emph{exception}, is at least $\mu_z$; likewise in $w$, with the minimum $\mu_w$. So if two of the six orders differ, the larger is the exception and the other five are equal to the smaller. If $c_A,c_B\ge1$, then $\zeta\ge1$ and $p^z_A,p^z_B,p^z_c\ge1$. \item $t^z_X\ge\min(p^z_X,\zeta-3r)$, with equality if $p^z_X\ne\zeta-3r$. If $r<0$ and $p^z_X=\mu_z$, then $t^z_X=\mu_z$. The same holds in $w$. \item In the modes F and Z1, $d_X+t^z_X\ge-\bar x_X$, with equality in mode F in the models of $\Vtwo$ and $\VN$. In mode Zs, $d_X+t^z_X=-x^0_X$. \item In mode Z1, $3\bar x_X<2+f_X$; in mode Zs, $3x^0_X<2+f_X$. Moreover $3r<2+\zeta$ and $3r<2+\omega$. \item In the modes Z1 and Zs, $c_X+d_X\ge1+f_X=2\bar x_X$. If $r>0$, the minimum of $c_A+d_A$, $c_B+d_B$ and $0$ is attained at least twice. If $r<0$, the minimum of $2r$, $c_A+d_A$ and $c_B+d_B$ is attained at least twice, and it is at most $2r$. \end{enumerate} \end{lemma} \begin{proof} (a) If two of the six orders exceeded the minimum, they would form, with an order equal to the minimum, a triple whose minimum is attained once, against (P). If $c_A,c_B\ge1$ and $\mu_z<1$, both $c_A$ and $c_B$ would be exceptions. (b) is (E): the two smallest of $p^z_X$, $\zeta-3r$ and $t^z_X$ are equal. If $r<0$, then $\zeta-3r>\zeta\ge\mu_z$. (c) follows from (S); for example, in mode F, $d_X+t^z_X\ge d_X+c_X-3\bar x_X=-\bar x_X$. (d) restates $f_X<1$, $3s_X+f_X<4$, $\zeta+3\omega<4$ and $3\zeta+\omega<4$. (e) The inequality holds as $c_X\ge1$. The rest is (I), whose minimum is at most $0<2r$ if $r>0$, and at most $2r<0$ if $r<0$. \end{proof} \subsection{The pairs of modes}\label{ss:d5pairs} \begin{proposition}\label{prop:d5empty} The models of the pairs of modes \textup{(F,\,Zs)}, \textup{(Z1,\,Ws)} and \textup{(Zs,\,Ws)} are empty. \end{proposition} \begin{proof} Let $B$ be in mode Ws$_{t_0}$, with the far order $g:=c_B<1$, so that $d_B\ge1$, $s_B>1$, $x^0_B=\frac12(s_B+g)$, $t^z_B=g-3x^0_B$ and $t^w_B=s_B-3x^0_B$. Let $A$ be in mode Z1 or Zs. By Lemma~\ref{lem:d5rules}(a), $c_B<1\le c_A$ gives $\zeta=p^z_A=p^z_B=p^z_c=g$, and $d_A<1\le d_B$ gives $\omega=p^w_A=p^w_B=p^w_c=f_A$. So $2r=f_A+g$. In $z$, relation (E) for $B$ says that the maximum of $0$, $r$, $x^0_B$ is attained twice. In $w$, since $t^w_B-f_A=-x^0_B-2r$, it says that the maximum of $0$, $3r$, $x^0_B+2r$ is attained twice. If $r>0$, the first statement gives $x^0_B=r$, hence $s_B=f_A<1$. If $r<0$, it gives $x^0_B=0$, and then $0$ is the only maximum in the second. If $r=0$, it gives $x^0_B\le0$, hence $s_B\le-g=f_A<1$. Each case contradicts $s_B>1$. Now let $A$ be in mode F and $B$ in mode Zs$_{t_0}$. By Lemma~\ref{lem:d5rules}(a), $c_A<1\le c_B$ gives $\zeta=c_A=p^z_A=p^z_B=p^z_c=:\kappa<1$. Thus $\omega=2r-\kappa$. If $r>0$, then $\zeta-3r<\kappa$, and Lemma~\ref{lem:d5rules}(b) gives $t^z_A=t^z_B=\kappa-3r$. By (S), $t^z_A\ge\kappa-3\bar x_A$. Hence $\bar x_A\ge r$, that is, $d_A\ge\omega$. And $t^z_B=-\frac12(s_B+3f_B)=\kappa-3r$ with $s_B>1$ gives $f_B<2r-\frac13(2\kappa+1)<\omega$. So $\omega$ and $d_A$ both exceed $f_B$ and both are exceptions in $w$. If $r<0$, then $\zeta-3r>\kappa$, and $t^z_A=t^z_B=\kappa$. For $A$ this gives $\bar x_A\ge0$. Thus $d_A\ge-\kappa>\omega$. So $d_A$ is the exception in $w$ and $f_B=\omega=p^w_B$. For $B$ it gives $f_B=-\frac13(2\kappa+s_B)$. As $\omega-3r>p^w_B$, Lemma~\ref{lem:d5rules}(b) gives $t^w_B=p^w_B=f_B$. Then $x^0_B=0$, $s_B=-f_B$ and $f_B=-\kappa$, whereas $f_B=\omega<-\kappa$. If $r=0$, then $t^z_B\ge\kappa$, so $s_B+3f_B\le-2\kappa$ and $f_B<-\frac13(2\kappa+1)<-\kappa=\omega$. So $\omega$ is the exception in $w$ and $d_A=f_B=p^w_A<\omega$. Hence $t^w_A=f_B$, while (S) gives $t^w_A\ge d_A-3\bar x_A=-\frac12(3\kappa+f_B)$, that is, $f_B\ge-\kappa$. \end{proof} \begin{proposition}\label{prop:d5zero} At every point of the model of $\Vtwo^{\mathrm w}$ for each of the following pairs of modes, $r=0$, and: \begin{enumerate}[label=\textup{(\alph*)}] \item \textup{(Z1,\,W1)}: $\bar x_A+\bar x_B=1$; \item \textup{(F,\,F)}: $\bar x_A,\bar x_B\ge0$ and $\bar x_A+\bar x_B<1$; in the models of $\Vtwo$ and $\VN$, $\bar x_A=\bar x_B=0$; \item \textup{(F,\,Z1)}: $\bar x_A,\bar x_B\ge0$ and $\bar x_A+\bar x_B<1$; in the model of $\Vtwo$, $\bar x_A=0$; \item \textup{(Z1,\,Z1)}: one of $f_A$, $f_B$ is equal to $\mu_w$, and $\mu_w\le\omega\le-1$; so $\bar x_A+\bar x_B<1$. \end{enumerate} \end{proposition} \begin{proof} (a) Let $g=c_B<1$ be the far order of $B$. By Lemma~\ref{lem:d5rules}(a), $\zeta=g$ and $\omega=f_A$. Thus $2r=f_A+g$. Since $c_A+d_A\ge1+f_A>2r$ and $c_B+d_B\ge1+g>2r$, the minimum in (I) is attained by $2r$ and $0$. So $r=0$ and $\bar x_A+\bar x_B=1+\frac12(f_A+g)=1$. (b) Suppose that $r>0$. Since the minimum in (V) is at most $-r<0$, it is attained by $d_X+t^z_X$ for some $X$, and Lemma~\ref{lem:d5rules}(c) gives $\bar x_X\ge r>0$. By Lemma~\ref{lem:d5rules}(e), the other body $Y$ has $c_Y+d_Y=0$. The minimum of $c_A$, $c_B$, $\zeta$ is attained twice. So $c_Y\ge\min(c_X,\zeta)$, and likewise $d_Y\ge\min(d_X,\omega)$; hence $0=c_Y+d_Y\ge\min(c_X,\zeta)+\min(d_X,\omega)$. As $c_X+d_X>0$ and $\zeta+\omega>0$, either $c_X\le\zeta$ and $\omega\le d_X$, or $\zeta\le c_X$ and $d_X\le\omega$. In the first case also $\omega\le d_Y$. Thus $\omega=\mu_w$. Then $p^w_Y\ge\omega>\omega-3r$ and $t^w_Y\ge d_Y-3\bar x_Y=d_Y\ge\omega$ by (S). Hence the minimum $\omega-3r$ in (E) is attained once. The second case is the same in $z$. Suppose that $r<0$. By Lemma~\ref{lem:d5rules}(e), $c_X+d_X\le2r$ for some $X$. Then $\bar x_X<0$ and $d_X+t^z_X\ge-\bar x_X>0$. As $-r>0$, the minimum in (V) is $0$, attained also by $d_Y+t^z_Y\ge-\bar x_Y$. Hence $\bar x_Y\ge0$. Then $c_Y$ and $d_Y$ are not both minimal in their colours, since otherwise $2\bar x_Y\le c_X+d_X<0$. Say that $c_Y$ is the exception in $z$. Then $p^z_X=c_X=\mu_z$. Lemma~\ref{lem:d5rules}(b) gives $t^z_X=c_X$, against $t^z_X\ge c_X-3\bar x_X>c_X$. The case of $d_Y$ is the same in $w$. So $r=0$. If $\bar x_A<0$, then $c_A+d_A<\zeta+\omega$; say $c_A<\zeta$. Then $\zeta$ is the exception in $z$ and $p^z_A=c_A<\zeta-3r$. So $t^z_A=c_A$, against $t^z_A\ge c_A-3\bar x_A>c_A$. Hence $\bar x_A,\bar x_B\ge0$. If $\zeta$ were the exception in $z$, then $c_A=c_B=\mu_z$ and $\omega=-\zeta<-\mu_z\le d_A,d_B$, which would give two exceptions in $w$. So $\zeta=\mu_z$, likewise $\omega=\mu_w$, and $\mu_z+\mu_w=0$. One of $c_A$, $c_B$ is equal to $\mu_z$ and one of $d_A$, $d_B$ to $\mu_w$. Hence $\bar x_A+\bar x_B$ is half the sum of the two other orders among $c_A$, $d_A$, $c_B$, $d_B$, which are less than $1$. In the models of $\Vtwo$ and $\VN$, (S) and Lemma~\ref{lem:d5rules}(b) give $c_A-3\bar x_A=t^z_A\ge\mu_z$ and $d_A-3\bar x_A\ge\mu_w$. Adding, we obtain $-4\bar x_A\ge0$. So $\bar x_A=0$ and likewise $\bar x_B=0$. (c) By Lemma~\ref{lem:d5rules}(a), $c_A<1\le c_B$ gives $\zeta=c_A=p^z_A=p^z_B=p^z_c=:\kappa<1$. If $r>0$, Lemma~\ref{lem:d5rules}(b) gives $t^z_A=t^z_B=\kappa-3r$. For $A$, (S) then gives $\bar x_A\ge r>0$. Thus Lemma~\ref{lem:d5rules}(e) gives $c_B+f_B=0$ and $f_B\le-1$. For $B$, (S) gives $\kappa-3r\ge1-3\bar x_B$, that is, $f_B\ge2r-\frac13(2\kappa+1)>-1$. If $r<0$, then $t^z_A=t^z_B=\kappa$, so $\bar x_A\ge0$ and $f_B\ge-\frac13(2\kappa+1)>-1$. Then $c_A+d_A>2r$ and $c_B+f_B>0>2r$, against Lemma~\ref{lem:d5rules}(e). So $r=0$ and $\omega=-\kappa$. As in (b), $\bar x_A\ge0$. So $\omega$ is not the exception in $w$, since otherwise $d_A=\mu_w<\omega$ and $\bar x_A<0$. Hence $\omega=\mu_w=-\kappa>-1$. Then $f_B\ge-\kappa$ gives $\bar x_B>0$. One of $d_A$, $f_B$ is equal to $-\kappa$. If $d_A=-\kappa$, then $\bar x_A=0$ and $\bar x_B<1$. If $f_B=-\kappa$, then $\bar x_A+\bar x_B=\frac12(\kappa+d_A)+\frac12(1-\kappa)=\frac12(1+d_A)<1$. In the model of $\Vtwo$, (S) and Lemma~\ref{lem:d5rules}(b) give $\kappa-3\bar x_A=t^z_A\ge\kappa$, so $\bar x_A=0$. (d) By Lemma~\ref{lem:d5rules}(a), $\zeta\ge1$. Suppose that $r>0$. By Lemma~\ref{lem:d5rules}(e), $c_X+f_X\le0$ for some $X$. Then $f_X\le-1$ and $\bar x_X\le0$. As in (b), (V) gives $d_Y+t^z_Y\le-r$ for some $Y$. Hence $\bar x_Y\ge r>0$. Thus $Y\ne X$ and $f_Xf_X$. Then $p^w_X=f_X=\mu_w$. Lemma~\ref{lem:d5rules}(b) gives $t^w_X=f_X$, against $t^w_X\ge f_X-3\bar x_X>f_X$. So $r=0$ and $\omega=-\zeta\le-1$. By Lemma~\ref{lem:d5rules}(a), one of $f_A$, $f_B$ is equal to $\mu_w\le\omega$, and the other is less than $1$. \end{proof} \begin{proposition}\label{prop:d5ss} Let both outer bodies be in mode Zs. At every point of the model of $\Vtwo^{\mathrm w}$, $r\ge0$ and $x^0_A,x^0_B\le r$. If $r>0$, then $y$ lies in the triangle $\mathcal T=\conv\{(0,-\frac12,0,-\frac12,0),\,0,\,(\frac13,0,\frac13,0,\frac13)\}$. If $r=0$, then $y$ lies in the polytope $\mathcal R$ of the points with $y_5=0$, $y_2\le y_1\le0$, $y_4\le y_3\le0$, $y_2,y_4\ge-\frac12$ and \[ 3y_4\le y_1+2y_2\le2y_4,\qquad 3y_2\le y_3+2y_4\le2y_2 . \] \end{proposition} \begin{proof} For a body $X$ in mode Zs, $t^z_X=s_X-3x^0_X=1-x^0_X-2\bar x_X$ and $t^w_X=f_X-3x^0_X$. By Lemma~\ref{lem:d5rules}(c), relation (V) says that the maximum of $x^0_A$, $x^0_B$, $r$ and $0$ is attained at least twice. By Lemma~\ref{lem:d5rules}(a), $\zeta\ge1$ and $p^z_A,p^z_B,p^z_c\ge1$. \emph{$r\ge0$.} Suppose that $r<0$. If $p^w_A=p^w_B=\mu_w$, Lemma~\ref{lem:d5rules}(b) gives $t^w_X=\mu_w\le f_X$, so $3x^0_X=f_X-\mu_w\ge0$ for both $X$. If moreover $f_A=f_B=\mu_w$, then $x^0_X=0$ and $t^z_X=s_X=-f_X=-\mu_w$. One of $p^z_A$, $p^z_B$ is equal to $\mu_z$. Hence $t^z_X=\mu_z$ for that $X$, and $2r=\zeta+\omega\ge\mu_z+\mu_w=0$, against $r<0$. If instead $f_A>\mu_w$, then $f_B=\mu_w$ and $x^0_B=0\mu_w$. If $p^w_A>\mu_w$, then $f_A=f_B=\omega=p^w_B=\mu_w$ and $t^w_B=f_B$, so $x^0_B=0$ and $t^z_B=s_B=-\omega>\zeta\ge\mu_z$. By Lemma~\ref{lem:d5rules}(b), $p^z_B\ne\mu_z$, that is, the position of $B$ is the exception in $z$, and $c_A=\zeta=p^z_A=\mu_z$. Then $t^z_A=\zeta$. Thus $x^0_A=-\zeta-f_A=-2r>0$, and the maximum is attained once. The case $p^w_B>\mu_w$ is the same. \emph{$x^0_X\le r$.} Suppose that $x^0_A>r$. As $r\ge0$, the maximum of $x^0_A$, $x^0_B$, $r$, $0$ exceeds $r$ and $0$. Hence $x^0_A=x^0_B=:q>r$. If $t^w_X<\omega-3r$ for some $X$, Lemma~\ref{lem:d5rules}(b) gives $p^w_X=t^w_X=f_X-3q\omega$, which gives two exceptions in $w$. \emph{$r>0$.} Then $x^0_Y=r$ for some $Y$, by the maximum rule. In (E) for $Y$ in $w$, $t^w_Y=f_Y-3r$. We use Lemma~\ref{lem:d5rules}(b). If $f_Y>\omega$, then $t^w_Y>\omega-3r$ and $p^w_Y=\omega-3r<\omega$, while $f_Y$ is the exception and $p^w_Y=\omega$. If $f_Y<\omega$, then $p^w_Y=t^w_Yf_X-3r\ge\omega-3r$, hence $p^w_X=\omega-3r<\mu_w$. So $x^0_X=r$ and $f_X=\omega$ as for $Y$. Put $f=\omega$. If $1+f>0$, then $c_A+f>0$ and $c_B+f>0$, and the minimum $0$ in Lemma~\ref{lem:d5rules}(e) is attained once. So $f\le-1$, that is, $\bar x_A=\bar x_B=\frac12(1+f)\le0$. By Lemma~\ref{lem:d5rules}(d), $3r<2+\omega=1+2\bar x_A$. So $y=(r,\bar x_A,r,\bar x_A,r)$ with $0<3r<1+2\bar x_A$ and $\bar x_A\le0$, which lies in $\mathcal T$. \emph{$r=0$.} Put $L=\zeta=-\omega$, so that $L\ge1$, $\zeta-3r=L$ and $\omega-3r=-L$. As $x^0_X\le0$, $t^w_X\ge f_X$ and $t^z_X\ge s_X>1$. By Lemma~\ref{lem:d5rules}(a) and (b), in $z$ either (z1) $\zeta$ is the exception, $c_A=c_B=p^z_A=p^z_B=p^z_c=\mu_z0$, against $x^0_X\le0$. So $t^z_X\ge L$ for both $X$ in these cases. We show $y\in\mathcal R$. The inequalities $y_2\le y_1\le0$, $y_4\le y_3\le0$ and $y_2,y_4\ge-\frac12$ hold since $s_X>1$, $x^0_X\le0$ and $\bar x_X\ge-\frac12$. As $t^z_X=1-x^0_X-2\bar x_X$ and $f_Y=2\bar x_Y-1$, the other four inequalities say that \[ -f_Y\le t^z_X\le-\tfrac12(1+3f_Y)\qquad\text{for }\{X,Y\}=\{A,B\}. \] The lower bound holds with equality in (w1). In (w2), $t^z_A\ge L=-f_B$ and $t^z_B\ge L>-f_A$; (w3) is the same; and in (w4), $t^z_X\ge L=-f_Y$. For the upper bound, $s_X>1$ gives $t^z_X<-\frac12(1+3f_X)$, which suffices if $f_X\ge f_Y$. It remains to treat (w2) with $X=B$, and (w3) with $X=A$, which is the same. In (F) for $A$ in $w$, the terms are $f_A$, $-L+\beta$ and $t^w_A\ge f_A$. Hence $-L+\beta\ge f_A>-L$, that is, $\beta>0$. In (F) for $B$ in $w$, the terms are $f_B=-L$, $-L+\beta>-L$ and $t^w_B\ge-L$. So $t^w_B=-L=f_B$. Hence $x^0_B=0$, $s_B=L$ and $t^z_B=L$. Finally, $t^z_A\ge L$ gives $3f_A\le-2L-s_A<-2L-1$, that is, $t^z_B=L<-\frac12(1+3f_A)$. \end{proof} \begin{proposition}\label{prop:d5zs} Let $A$ be in mode Z1 and $B$ in mode Zs. At every point of the model of $\Vtwo^{\mathrm w}$, $y$ lies in $\mathcal S^0$ if $r=0$, in $\mathcal S^+$ if $r>0$, and in $\mathcal S^-$ if $r<0$, where \begin{align*} \mathcal S^0&=\{y:\ y_1=y_2,\ y_5=0,\ y_4\le y_3\le0,\ -\tfrac12\le y_4\le y_1\le1\},\\ \mathcal S^+&=\{y:\ y_1=y_2,\ y_3=y_5\ge0,\ -\tfrac12\le y_4\le0,\ 3y_5\le y_1+2y_4,\ y_1\le1\},\\ \mathcal S^-&=\{y:\ y_1=y_2,\ y_3=0,\ y_4\le y_5\le0,\ y_1+2y_4\ge0,\ y_1\le1\}. \end{align*} Moreover $x^0_B\le\max(r,0)$. If $r>0$, then $f_B=\omega\le-1$ and $t^w_A=\omega-3r$. If $r=0$, then $f_B\le f_A$. \end{proposition} \begin{proof} Here $y=(\bar x_A,\bar x_A,x^0_B,\bar x_B,r)$, with $\bar x_A=\frac12(1+f_A)<1$, $\bar x_B\ge-\frac12$ and $x^0_B>\bar x_B$. By Lemma~\ref{lem:d5rules}(a), $\zeta\ge1$ and $p^z_A,p^z_B,p^z_c\ge1$. By Lemma~\ref{lem:d5rules}(c), (V) says that the minimum of $f_A+t^z_A$, $-x^0_B$, $-r$ and $0$ is attained at least twice, and $f_A+t^z_A\ge-\bar x_A$. \emph{Step 1: $x^0_B\le\max(r,0)$.} Suppose not. Then the minimum in (V) is attained by $-x^0_B$ and $f_A+t^z_A$. Hence $\bar x_A\ge x^0_B>0$ and $f_A>-1$. If $t^w_B<\omega-3r$, Lemma~\ref{lem:d5rules}(b) gives $p^w_B=t^w_B\omega$. Thus again $f_B$ is the exception and $f_A=\omega$. If $r\le0$, then $\zeta=2r-\omega<1$ as before. If $r>0$, then $\zeta-3r=-r-\omega>-x^0_B-\omega=t^z_A$, and Lemma~\ref{lem:d5rules}(b) gives $p^z_A=t^z_A=-x^0_B-\omega$. As $p^z_A\ge1$, we obtain $\omega<-1$, against $\omega=f_A>-1$. \emph{Step 2: if $r>0$, then $x^0_B=r$.} If $x^0_B0$. By Lemma~\ref{lem:d5rules}(e), $c_B+f_B=0$. Thus $f_B\le-10$, then $f_B=\omega\le-1$, $t^w_A=\omega-3r$ and $3r\le\bar x_A+2\bar x_B$.} As $x^0_B=r$, (E) for $B$ in $w$ gives $f_B=\omega=\mu_w$, as in the proof of Proposition~\ref{prop:d5ss}. If $c_B+f_B>0$, Lemma~\ref{lem:d5rules}(e) gives $c_A+f_A=0$, so $f_B=\mu_w\le f_A\le-1$; otherwise $f_B\le-c_B\le-1$. In (F) for $B$ in $w$, the terms are $\omega$, $\omega+\beta$ and $t^w_B=\omega-3r$. Hence $\beta=-3r$. In (F) for $A$ in $w$, the terms are $f_A\ge\omega>\omega-3r=\omega+\beta$ and $t^w_A$. Thus $t^w_A=\omega-3r$. By (S), $t^w_A\ge f_A-3\bar x_A=-\frac12(3+f_A)$. Hence $6r\le3+f_A+2f_B$, which is $3r\le\bar x_A+2\bar x_B$. \emph{Step 4: if $r<0$, then $x^0_B=0$, $\bar x_B\le r$ and $\bar x_A+2\bar x_B\ge0$.} Since $x^0_B\le0$ by step 1, $t^w_B\ge f_B$. If $p^w_B=\mu_w$, Lemma~\ref{lem:d5rules}(b) gives $t^w_B=\mu_w\le f_B\le t^w_B$, so $x^0_B=0$ and $f_B=\mu_w$. If $p^w_B>\mu_w$, then $f_A=f_B=\omega=p^w_A=\mu_w$, and Lemma~\ref{lem:d5rules}(b) gives $t^w_A=f_A$. By (S), $f_A\ge-\frac12(3+f_A)$. Thus $\omega=f_A\ge-1$ and $\zeta=2r-\omega<1$, which is impossible. So $x^0_B=0$ and $f_B=\mu_w$. Next, $c_B+f_B\le2r$. Otherwise, by Lemma~\ref{lem:d5rules}(e), $c_A+f_A=2r$. Hence $f_A<-1$, $\bar x_A<0$ and $t^w_A\ge f_A-3\bar x_A>f_A$. Then $p^w_A\ne\mu_w$ by Lemma~\ref{lem:d5rules}(b). So $f_A=f_B=\omega$. Now $c_B>c_A$ makes $c_B$ the exception in $z$. Hence $p^z_B=\mu_z=\zeta$, and Lemma~\ref{lem:d5rules}(b) gives $t^z_B=\zeta$. As $x^0_B=0$, this gives $s_B=t^z_B=\zeta$ and $x^0_B=\frac12(\zeta+\omega)=r<0$, a contradiction. So $f_B\le2r-c_B\le2r-1$, that is, $\bar x_B\le r$. Finally, if $p^w_A=\mu_w$, then Lemma~\ref{lem:d5rules}(b) gives $t^w_A=\mu_w=f_B$, and (S) gives $f_B\ge-\frac12(3+f_A)$, that is, $\bar x_A+2\bar x_B\ge0$. If $p^w_A>\mu_w$, then $f_A=f_B=\omega$, and $s_B=-f_B$ gives $t^z_B=s_B=-\omega=\zeta-2r$. As $\zeta-2r<\zeta-3r$, (E) for $B$ in $z$ and Lemma~\ref{lem:d5rules}(b) give $p^z_B=t^z_B>\zeta$. So the position of $B$ is the exception in $z$ and $p^z_A=\zeta=\mu_z$. Then $t^z_A=\zeta$. By (S), $\zeta\ge-\frac12(1+3\omega)$, that is, $\omega\ge-1-4r$. But $\omega=f_B\le2r-1<-1-4r$. \emph{Step 5: if $r=0$, then $f_B\le f_A$.} Otherwise $f_A=\omega=\mu_w$. In (F) for $B$ in $w$, the terms are $f_B$, $\omega+\beta$ and $t^w_B\ge f_B$. Hence $\beta\ge f_B-\omega>0$. In (F) for $A$ in $w$, the terms are $\omega$, $\omega+\beta>\omega$ and $t^w_A$. Thus $t^w_A=\omega=f_A$, and (S) gives $f_A\ge-1$. Then $\bar x_B>\bar x_A\ge0$. Hence $x^0_B>\bar x_B>0$, against step 1. Steps 1--5, with $\bar x_A<1$, $\bar x_B\ge-\frac12$ and $x^0_B>\bar x_B$, give the three cases. \end{proof} \begin{proposition}\label{prop:d5VN} \begin{enumerate}[label=\textup{(\alph*)}] \item The models of $\VN$ for \textup{(Z1,\,W1)}, \textup{(F,\,Z1)} and \textup{(Z1,\,Zs)} are empty, and so is the model of Theorem~C0 of the supplementary material for \textup{(Z1,\,Zs$_{t_0}$)} in the centred case. \item At every point of the model of $\VN$ for \textup{(Z1,\,Z1)}, $\bar x_A=\bar x_B\le0$. \end{enumerate} \end{proposition} \begin{proof} Since these models contain the relations of the model of $\Vtwo$, Propositions~\ref{prop:d5zero} and~\ref{prop:d5zs} apply to them. By (R), a body $X$ in mode Z1 has at most one event. For each of the other quantities of (R) the opposite inequality holds. With (S), (Q) and (T), this means that $c_X=1$, that $t^z_X$ and $t^w_X$ take their lower bounds $1-3\bar x_X$ and $f_X-3\bar x_X$, that $\tau_X=-\bar x_X$ and, if $Y$ is in mode Zs, that $p^z_A,p^z_B,p^z_c\le1$ and (in the model of $\VN$) that $\tilde t_X=-\bar x_X$, except for the event. (a) \emph{(Z1,\,W1).} By Proposition~\ref{prop:d5zero}(a) and its proof, $r=0$, $\zeta=g:=c_B$, $\omega=f_A$ and $f_A+g=0$. Lemma~\ref{lem:d5rules}(b) gives $t^z_B\ge g$ and $t^w_B\ge f_A$. If $t^z_B$ took its lower bound $-\frac12(3+g)$, then $g\le-1$. If $t^w_B$ took its lower bound $-\frac12(1+3g)$, then $-\frac12(1+3g)\ge-g$, and again $g\le-1$. As $g=-f_A>-1$, both are events of $B$, against (R). \emph{(F,\,Z1).} By Proposition~\ref{prop:d5zero}(c) and its proof, $r=0$, $\bar x_A=0$, $\zeta=c_A=:\kappa<1$, $\omega=\mu_w=-\kappa$ and $f_B\ge-\kappa$. Lemma~\ref{lem:d5rules}(b) gives $t^z_B\ge\kappa$ and $t^w_B\ge-\kappa$. If $t^z_B=-\frac12(1+3f_B)$, then $f_B\le-\frac13(2\kappa+1)<-\kappa$. So $t^z_B$ is the event of $B$. Then $t^w_B=-\frac12(3+f_B)\ge-\kappa$ gives $f_B\le2\kappa-3<-1$, against $f_B\ge-\kappa>-1$. \emph{(Z1,\,Zs).} Let $A$ be in mode Z1 and $B$ in mode Zs$_{t_0}$, in the model of $\VN$ or in that of the centred case. By Proposition~\ref{prop:d5zs}, $y$ lies in $\mathcal S^0$, $\mathcal S^+$ or $\mathcal S^-$. We say that \emph{det holds} if $p^z_A,p^z_B,p^z_c\le1$; it fails only if this is the event of $A$. As these positions are at least $1$, det means that they are equal to $1=\mu_z$. If det fails, then $c_A=1=\mu_z$. If $r=0$, then $x^0_B\le0$ and $t^z_B\ge s_B>1$. If det holds, (E) for $B$ in $z$ has the terms $p^z_B=1$, $\zeta\ge1$ and $t^z_B>1$, so $\zeta=1$. If det fails, $\zeta>1$ would make $\zeta$ the exception in $z$, and Lemma~\ref{lem:d5rules}(b) would give $t^z_B=p^z_B=1$; so again $\zeta=1$ and $\omega=-1$. If $\omega$ were the exception in $w$, then $p^w_A=f_A<-1$, and Lemma~\ref{lem:d5rules}(b) would give $t^w_A=f_A$, against $t^w_A\ge-\frac12(3+f_A)>f_A$. So $\omega=\mu_w=-1$ and one of $f_A$, $f_B$ is equal to $-1$. As $f_B\le f_A$, this gives $f_B=-1$. Then $\bar x_B=01$. If det holds, then $p^z_B=1=\mu_z$, and Lemma~\ref{lem:d5rules}(b) gives $t^z_B=1$, against $t^z_B>1$. If det fails, $p^z_B=\mu_z$ would give the same, so the position of $B$ is the exception in $z$ and $c_A=\zeta=p^z_A=1$. Then $t^z_A=1$ by Lemma~\ref{lem:d5rules}(b). As $t^z_A$ takes its lower bound $-\frac12(1+3f_A)$, $f_A=-1$, that is, $\bar x_A=0$. Then $\bar x_A+2\bar x_B\ge0$ gives $\bar x_B\ge0>r$, against $\bar x_B\le r$. If $r>0$, then $x^0_B=r$ and $f_B=\omega\le-1$, so $s_B=2r-\omega=\zeta>1$ and $t^z_B=\zeta-3r$. In the centred case $c_B>1$. If det holds, $\zeta>1=\mu_z$ makes $\zeta$ the exception in $z$, so $c_B=1$. If det fails, $c_A=1=\mu_z$ makes $c_B$ the exception, so $\zeta=1$. Both are impossible. In the model of $\VN$, suppose first that det holds and that $t^z_A$ and $t^w_A$ take their lower bounds. By (E) for $B$ in $z$, with the terms $p^z_B=1$ and $t^z_B=\zeta-3r$, we have $\zeta-3r\le1$. The equality $t^w_A=\omega-3r=-\frac12(3+f_A)$ gives $f_A=6r-3-2\omega$. If $\zeta-3r<1=p^z_A$, Lemma~\ref{lem:d5rules}(b) gives $t^z_A=\zeta-3r=-r-\omega$, and $t^z_A=-\frac12(1+3f_A)$ then gives $\omega=2r-1$ and hence $r=\frac12(1+\omega)\le0$. If $\zeta-3r=1$, then $\omega=-1-r$, $f_A=8r-1$, and $t^z_A\ge1$ gives $f_A\le-1$ and hence $r\le0$. Hence the event of $A$ is the failure of det, or $t^z_A$, or $t^w_A$. In each case $\tilde t_A=-\bar x_A$. In (T), $e_1-e_2=\frac12(3-2r-\omega)>0$, as $\omega\le-1$ and $r<1$. Hence $\tilde t_B=e_2=-\frac12(1+\omega)$. If $t^w_A$ takes its lower bound, then $\bar x_A=3r-1-\omega$ and $\tilde t_A=1+\omega-3r<\tilde t_B$, so (T) gives $p^w_c+1=\tilde t_A$, that is, $p^w_c=\omega-3r<\mu_w$, which is impossible. If $t^z_A$ takes its lower bound and the event is $t^w_A$, then det holds, and as above $t^z_A=\zeta-3r$ if $\zeta-3r<1$ and $t^z_A\ge1$ if $\zeta-3r=1$. In the first case $\bar x_A=\frac13(1+r+\omega)$. Then $3r\le\bar x_A+2\bar x_B$ gives $8r\le4+4\omega\le0$. In the second case $\bar x_A\le0$ and $\bar x_B=\frac12(1+\omega)=-\frac12r$. Then $3r\le\bar x_A+2\bar x_B$ gives $4r\le\bar x_A\le0$. Both contradict $r>0$. (b) By Proposition~\ref{prop:d5zero}(d), $r=0$, and one of $f_A$, $f_B$ is equal to $\mu_w\le-1$. By (Q) and (R), $\tau_A=\tau_B$ and $\tau_X\ge-\bar x_X$, with equality unless $\tau_X$ is the event of $X$. Suppose that $f_B-\bar x_A$. So $\tau_A$ is the event of $A$. Hence $c_A=1$ and $t^z_A$, $t^w_A$ take their lower bounds. By Lemma~\ref{lem:d5rules}(a), $f_B=\omega=-\zeta$ and $\mu_z=1$. Lemma~\ref{lem:d5rules}(b) gives $t^z_A\ge1$, that is, $f_A\le-1$. If $\zeta>1$, then $\zeta$ is the exception in $z$, and Lemma~\ref{lem:d5rules}(b) gives $t^z_A=t^z_B=1$. So $f_A=-1$. Then $t^z_B=1\ge-\frac12(1+3f_B)$ contradicts $f_B<-1$. If $\zeta=1$, then $f_B=-1