+\section{Merge}
+
+Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
+\gathbegin
+ C \hasparents \{ L, R \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ D \isin C \equiv
+ \begin{cases}
+ (D \isin L \land D \isin R) \lor D = C : & \true \\
+ (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
+ \text{otherwise} : & D \not\isin M
+ \end{cases}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Tip Merge }{
+ L \in \py \implies
+ \begin{cases}
+ R \in \py : & \baseof{R} \ge \baseof{L}
+ \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+ R \in \pn : & R \ge \baseof{L}
+ \land M = \baseof{L} \\
+ \text{otherwise} : & \false
+ \end{cases}
+}\]
+
+\subsection{No Replay}
+
+\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.
+
+\subsubsection{For $D \isin L \land D \isin R$:}
+$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK.
+
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \not\isin M$:}
+$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$. OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \isin M$:}
+$D \not\isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$. OK.
+
+$\qed$
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$,
+and calculate $\pendsof{C}{\pn}$. So we will consider some
+putative ancestor $A \in \pn$ and see whether $A \le C$.
+
+$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+But $C \in py$ and $A \in \pn$ so $A \neq C$.
+Thus $A \le L \lor A \le R$.
+
+By Unique Base of L and Transitive Ancestors,
+$A \le L \equiv A \le \baseof{L}$.
+
+\subsubsection{For $R \in \py$:}
+
+By Unique Base of $R$ and Transitive Ancestors,
+$A \le R \equiv A \le \baseof{R}$.
+
+But by Tip Merge condition on $\baseof{R}$,
+$A \le \baseof{L} \implies A \le \baseof{R}$, so
+$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} =
+\baseof{R}$.
+
+UP TO HERE
+
+By Tip Merge, $A \le $
+
+Let $S =
+ \begin{cases}
+ R \in \py : & \baseof{R} \\
+ R \in \pn : & R
+ \end{cases}$.
+Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
+
+Consider some $A \in \pn$. If $A \le S$ then $A \le C$.
+If $A \not\le S$ then
+
+Let $A \in \pends{C}{\pn}$.
+Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
+\pendsof{R,\pn}$.
+
+
+
+%$\pends{C,
+
+%%\subsubsection{For $R \in \py$:}
+%foo
+