\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,
-$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+$D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
+$D \isin L \equiv D \isin M$. So by $\merge$, $D \isin
C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
Thus $D \isin C \equiv D \isin \baseof{C}$. OK.