+\section{Anticommit}
+
+Given $L, R^+, R^-$ where
+$R^+ \in \pry, R^- = \baseof{R^+}$.
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \merge{C}{L}{R^+}{R^-}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+
+\subsection{Desired Contents}
+
+xxx need to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+
+\subsection{No Replay}
+
+By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
+so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$ and No Replay for
+Merge Results applies. $\qed$
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$.
+
+xxx tbd
+
+\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
+ \merge{C}{L}{M}{R}
+\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{Merge Results}
+
+As above.
+
+\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 $A \le C$.
+
+By Exact Ancestors for 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 $A \le C \equiv A \le L \lor A \le R$.
+
+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
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.
+That is, $\baseof{C} = \baseof{R}$.
+
+\subsubsection{For $R \in \pn$:}
+
+By Tip Merge condition on $R$,
+$A \le \baseof{L} \implies A \le R$, so
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.
+Thus $A \le C \equiv A \le R$.
+That is, $\baseof{C} = R$.
+
+$\qed$
+