\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}
so $L \in \py$ so $L \haspatch \p$. We will use the coherence and
patch inclusion of $C$ as just proved.
-Firstly we prove $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+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 $\nothaspatch$).
-xxx up to here
-
We will consider some $D$ and prove the Exclusive Tip Contents form.
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$. OK.
-So by definition of
-$\haspatch$, $D \isin C \equiv D \le C$. OK.
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
-\subsubsection{For $L \in \py, D \in \py, $:}
-$R \haspatch \p$ so
+$D \neq C$. 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 $L \in \py, D \not\in \py, R \in \py$:}
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
%D \in \py$:}