\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
\end{cases}
$
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \equiv D \isin R : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin M \equiv D \isin L : & D = C \lor D \isin R \\
+ D \isin M \equiv D \isin R : & D = C \lor D \isin L \\
+ \end{cases}
+$
+
\end{basedescript}
\newpage
\section{Invariants}
\text{otherwise} : & \false
\end{cases}
}\]
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
\[ \eqn{ Removal Merge Ends }{
X \not\haspatch \p \land
Y \haspatch \p \land
$\qed$
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
+R$. And $D \neq C$. So $D \not\isin C$. $\qed$
+
\subsection{Tip Contents}
+We need worry only about $C \in \py$.
+And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$. We will use the coherence and
+patch inclusion of $C$ as just proved.
+
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by coherence/inclusion $C \haspatch \p$ . If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$. So by coherence/inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
+
We will consider some $D$ and prove the Exclusive Tip Contents form.
-We use the Coherence of $C$ as just proved.
+
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$. OK.
+
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
+
+$D \neq C$. By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$. xxx up to here
+
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+
+%D \in \py$:}
+
+
xxx the coherence is not that useful ?
-\subsubsection{For $L \in \py, D \in \py$:}
+$L \haspatch \p$ by
xxx need to recheck this