-If $D = C$, trivial. For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
-
-\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
- \mergeof{C}{L}{R^+}{R^-}
-\end{gather}
-
-\subsection{Conditions}
-
-\[ \eqn{ Unique Tip }{
- \pendsof{L}{\pry} = \{ R^+ \}
-}\]
-\[ \eqn{ Currently Included }{
- L \haspatch \pry
-}\]
-\[ \eqn{ Not Self }{
- L \not\in \{ R^+ \}
-}\]
-
-\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{Desired Contents}
-
-\[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \]
-\proofstarts
-
-\subsubsection{For $D = C$:}
-
-Trivially $D \isin C$. OK.
-
-\subsubsection{For $D \neq C, D \not\le L$:}
-
-By No Replay $D \not\isin L$. Also $D \not\le R^-$ hence
-$D \not\isin R^-$. Thus $D \not\isin C$. OK.
-
-\subsubsection{For $D \neq C, D \le L, D \in \pry$:}
-
-By Currently Included, $D \isin L$.
-
-By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by
-by Unique Tip, $D \le R^+ \equiv D \le L$.
-So $D \isin R^+$.
-
-By Base Acyclic, $D \not\isin R^-$.
-
-Apply $\merge$: $D \not\isin C$. OK.
-
-\subsubsection{For $D \neq C, D \le L, D \notin \pry$:}
-
-By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
-
-Apply $\merge$: $D \isin C \equiv D \isin L$. OK.
-
-$\qed$
-
-\subsection{Unique Base}
-
-Need to consider only $C \in \py$, ie $L \in \py$.
-
-xxx tbd
-
-xxx need to finish anticommit
-
-\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
- \mergeof{C}{L}{M}{R}
-\end{gather}
-We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
-
-\subsection{Conditions}
-
-\[ \eqn{ Tip Merge }{
- L \in \py \implies
- \begin{cases}
- R \in \py : & \baseof{R} \ge \baseof{L}
- \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
- R \in \pn : & R \ge \baseof{L}
- \land M = \baseof{L} \\
- \text{otherwise} : & \false
- \end{cases}
-}\]
-\[ \eqn{ Removal Merge Ends }{
- X \not\haspatch \p \land
- Y \haspatch \p \land
- M \haspatch \p
- \implies
- \pendsof{Y}{\py} = \pendsof{M}{\py}
-}\]
-\[ \eqn{ Addition Merge Ends }{
- X \not\haspatch \p \land
- Y \haspatch \p \land
- M \nothaspatch \p
- \implies \left[
- \bigforall_{E \in \pendsof{X}{\py}} E \le Y
- \right]
-}\]
-
-\subsection{No Replay}
-
-See No Replay for Merge Results.
-
-\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, $\baseof{C} = R$.
-
-$\qed$
-
-\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. $C \nothaspatch \p$.