We will show for each of
various cases that
if $M \haspatch \p$, $D \not\isin C$,
-whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
+whereas if $M \nothaspatch \p$, $D \isin C \equiv D \le C$.
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.
Consider $D = C$: Thus $C \in \py, L \in \py$.
-By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
+By Tip Own Contents, $L \haspatch \p$ so $L \neq X$,
therefore we must have $L=Y$, $R=X$.
-By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
+Conversely $R \not\in \py$
+so by Tip Merge $M = \baseof{L}$. Thus $M \in \pn$ so
by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
and $D \le C$. OK.