\section{Anticommit}
Given $L, R^+, R^-$ where
-$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.
+$R^+ \in \pry, R^- = \baseof{R^+}$.
Construct $C$ which has $\pr$ removed.
Used for removing a branch dependency.
\gathbegin
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
}\]
-\[ \eqn{ Correct Base }{
- \baseof{R^+} = R^-
-}\]
\[ \eqn{ Currently Included }{
L \haspatch \pry
}\]
+\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 want to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+xxx tbd
\section{Merge}