+\subsection{Coherence and Patch Inclusion}
+
+Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
+This involves considering $D \in \py$.
+
+\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
+$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
+\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch). So $D \neq C$.
+Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+
+\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
+$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
+(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
+
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
+
+For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
+ \equiv D \isin L \lor D \isin R$.
+(Likewise $D \le C \equiv D \le X \lor D \le Y$.)
+
+Consider $D \neq C, D \isin X \land D \isin Y$:
+By $\merge$, $D \isin C$. Also $D \le X$
+so $D \le C$. OK for $C \haspatch \p$.
+
+Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
+By $\merge$, $D \not\isin C$.
+And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
+OK for $C \haspatch \p$.
+
+Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
+$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
+Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$.
+OK for $C \haspatch \p$.
+
+So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$C \haspatch \p \equiv M \nothaspatch \p$.
+
+\proofstarts
+
+One of the Merge Ends conditions applies.
+Recall that we are considering $D \in \py$.
+$D \isin Y \equiv D \le Y$. $D \not\isin X$.
+We will show for each of
+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
+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 Addition 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 $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
+and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
+Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \not\isin Y$:
+By $\merge$, $D \not\isin C$. OK.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
+R$. And $D \neq C$. So $D \not\isin C$. $\qed$
+
+\subsection{Tip Contents}
+
+We need worry only about $C \in \py$.
+And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
+of $C$, and its Coherence and Patch Inclusion, as just proved.
+
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by Coherence/Inclusion $C \haspatch \p$ . If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$. So by Coherence/Inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
+
+We will consider an arbitrary commit $D$
+and prove the Exclusive Tip Contents form.
+
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$. OK.
+
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
+
+$D \neq C$. By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+$D \neq C$.
+
+By Tip Contents
+$D \isin L \equiv D \isin \baseof{L}$ and
+$D \isin R \equiv D \isin \baseof{R}$.
+
+If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
+Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
+$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
+$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
+
+So $D \isin M \equiv D \isin L$ and by $\merge$,
+$D \isin C \equiv D \isin R$. But from Unique Base,
+$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$. OK.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+By Foreign Inclusion of $L, M, R$:
+$D \isin L \equiv D \le L$;
+$D \isin M \equiv D \le M$;
+$D \isin R \equiv D \le R$.
+
+\subsubsection{For $D = C$:}
+
+$D \isin C$ and $D \le C$. OK.
+
+\subsubsection{For $D \neq C, D \isin M$:}
+
+Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
+R$. So by $\merge$, $D \isin C$. And $D \le C$. OK.
+
+\subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
+
+By $\merge$, $D \isin C$.
+And $D \isin X$ means $D \le X$ so $D \le C$.
+OK.
+
+\subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
+
+By $\merge$, $D \not\isin C$.
+And $D \not\le L, D \not\le R$ so $D \not\le C$.
+OK
+
+$\qed$
+