\end{cases}
$$
\proofstarts
-~
Firstly, if $L \haspatch \p$, $\exists_{F \in \py} F \le L$
and this $F$ is also $\le C$
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
+OK.
\subsubsection{For $L \nothaspatch \p$:}
Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
-Now by contents of $L$, $D \notin L$, so $D \notin C$.
+Now by $\nothaspatch$, $D \not\isin L$ so $D \not\isin C$.
OK.
$\qed$
+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies. $\qed$
+
\subsection{Foreign Inclusion:}
Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Contents:}
-Only relevant if $\patchof{C} = \bot$, and in that case Totally
+Only relevant if $\isforeign{C}$, and in that case Totally
Foreign Contents applies. $\qed$