-$D \isin Y \equiv D \le Y$. $D \not\isin X$. Recall that we
-are considering $D \in \py$.
-
-Consider $D = C$. Thus $C \in \py, L \in \py$.
-But $X \not\haspatch \p$ means xxx wip
-But $X \not\haspatch \p$ means $D \not\in X$,
-
-so we have $L = Y, R =
-X$. Thus $R \not\haspatch \p$ and by Tip Self Inpatch $R \not\in
-\py$. Thus by Tip Merge $R \in \pn$ and $M = \baseof{L}$.
-So by Base Acyclic, $M \nothaspatch \py$. Thus we are expecting
-$C \haspatch \py$. And indeed $D \isin C$ and $D \le C$. OK.
+Consider $D = C$. Thus $C \in \py, L \in \py$, and by Tip
+Self Inpatch $L \haspatch \p$, so $L=Y, R=X$. By Tip Merge,
+$M=\baseof{L}$. So by Base Acyclic $D \not\isin M$, i.e.
+$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK.