\subsubsection{For $\p = \pq$:}
By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
-By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \haspatch \pq$.
+By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \zhaspatch \pq$.
+And we can set $F = C$ giving $F \in \pqy \land F \le C$ so $C
+\haspatch \pq$.
\subsubsection{For $\p \neq \pq$:}