$A \le C$ so $D \le C$.
}
+\[ \eqn{Simple Foreign Inclusion:}{
+ \left[
+ C \hasparents \{ L \}
+ \land
+ \bigforall_{D} D \isin C \equiv D \isin L \lor D = C
+ \right]
+ \implies
+ \bigforall_{D \text{ s.t. } \patchof{D} = \bot}
+ D \isin C \equiv D \le C
+}\]
+\proof{
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+If $D = C$, trivially true. For $D \neq C$,
+by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
+
\[ \eqn{Totally Foreign Contents:}{
\bigforall_{C \hasparents \set A}
\left[
\[ A \nothaspatch P \implies C \nothaspatch P \]
$\qed$
-\subsection{Foreign inclusion:}
+\subsection{Foreign Inclusion:}
-If $D = C$, trivial. For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
+Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Contents:}
\subsection{Coherence and Patch Inclusion}
Consider some $D \in \p$.
-$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$.
+$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$
+and $D \le B \equiv D \le L$.
Thus $L \haspatch \p \implies B \haspatch P$
and $L \nothaspatch \p \implies B \nothaspatch P$.
\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq B$
-so $D \isin B \equiv D \isin L$.
-By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
-And by Exact Ancestors $D \le L \equiv D \le B$.
-So $D \isin B \equiv D \le B$. $\qed$
+Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Contents}
\subsection{Unique Base}
-Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$.
+Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. $\qed$
\subsection{Tip Contents}
\subsection{Coherence and Patch Inclusion}
-Consider some $D \in \py$.
+$$
+\begin{cases}
+ \p = \pq \lor B \haspatch \p : & C \haspatch \p \\
+ \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+
+\proofstarts
+~ Consider some $D \in \py$.
\subsubsection{For $\p = \pq$:}
By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
By No Sneak, $D \le B \equiv D = C$. Thus $C \haspatch \pq$.
+\subsubsection{For $\p \neq \pq$:}
+
+$D \neq C$. So $D \isin C \equiv D \isin B$,
+and $D \le C \equiv D \le B$.
+
+$\qed$
+
xxx up to here
\section{Anticommit}