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
$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
\le C$. OK.
-xxx up to here
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
-\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+$D \neq C$. 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$:}