therefore we must have $L=Y$, $R=X$.
By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
-and $D \le C$, consistent with $C \haspatch \p$. OK.
+and $D \le C$. OK.
Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
$D \le Y$ so $D \le C$.