$\qed$
-\subsection{Coherence and patch inclusion}
+\subsection{Coherence and Patch Inclusion}
-Need to consider $D \in \py$
+$$
+\begin{cases}
+ L \haspatch \p : & C \haspatch \p \\
+ L \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+\proofstarts
+
+Firstly, if $L \haspatch \p$, $\exists_{F \in \py} F \le L$
+and this $F$ is also $\le C$
+so $C \zhaspatch \p \implies C \haspatch \p$.
+We will prove $\zhaspatch$
+
+We need to consider $D \in \py$.
\subsubsection{For $L \haspatch \p, D = C$:}
$ D \le C $.
Contents of $C$:
-$ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $.
+$ D \isin C \equiv \ldots \lor \true$. So $ D \zhaspatch C $.
+OK.
\subsubsection{For $L \haspatch \p, D \neq C$:}
Ancestors: $ D \le C \equiv D \le L $.
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
-By $\haspatch$ for $L$, $\exists_{F \in \py} F \le L$
-and this $F$ is also $\le C$. So $\haspatch$.
-
-So:
-\[ L \haspatch \p \implies C \haspatch \p \]
+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.
-So:
-\[ L \nothaspatch \p \implies C \nothaspatch \p \]
$\qed$
+\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. $\qed$