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.
+so $L \in \py$ so $L \haspatch \p$. We will use the unique base,
+and 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 $R \not\in \py$
$D \neq C$. 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
-
+$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
\subsubsection{For $D \not\in \py, R \in \py$:}
+xxx up to here
%D \in \py$:}