various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
(which suffices by definition of $\haspatch$ and $\nothaspatch$).
-Consider $D = C$. Thus $C \in \py, L \in \py$, and by Tip
+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.
+Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+$D \le Y$ so $D \le C$.
+$D \not\isin M$ so by $\merge$, $D \isin C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
+$D \not\le Y$. If $D \le X$ then
+$D \in \pancsof{X}{\py}$, so by Merge Ends and
+Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
+Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \isin Y$:
+$D \le Y$ so $D \le M$
+
+%anyway D \not\isin X
+% D \isin Y
+% D \le Y
+
+%bad case
+% D \not\le M
+% D \not\isin M
+% results
+% D \isin C wrong
+
+%ok case
+% D \le M
+% D \isin M
+% results
+% D \not\isin C OK
+
\end{document}