-Removal Merge Ends: For $\p = \pc$, $M \nothaspatch \p$; OK.
-For $\p \neq \pc$, by Tip Contents,
-$M \haspatch \p \equiv L \haspatch \p$, so we need only
-worry about $X = R, Y = L$; ie $L \haspatch \p$,
-$M = \baseof{L} \haspatch \p$.
-By Tip Contents for $L$, $D \le L \equiv D \le M$. OK.~~$\qed$