-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}
-
-\[ \eqn{ Tip Merge }{
- L \in \py \implies
- \begin{cases}
- R \in \py : & \baseof{R} \ge \baseof{L}
- \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
- R \in \pn : & M = \baseof{L} \\
- \text{otherwise} : & \false
- \end{cases}
-}\]
-\[ \eqn{ Merge Acyclic }{
- L \in \pn
- \implies
- R \nothaspatch \p
-}\]
-\[ \eqn{ Removal Merge Ends }{
- X \not\haspatch \p \land
- Y \haspatch \p \land
- M \haspatch \p
- \implies
- \pendsof{Y}{\py} = \pendsof{M}{\py}
-}\]
-\[ \eqn{ Addition Merge Ends }{
- X \not\haspatch \p \land
- Y \haspatch \p \land
- M \nothaspatch \p
- \implies \left[
- \bigforall_{E \in \pendsof{X}{\py}} E \le Y
- \right]
-}\]
-
-\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 $R$ and since $M \le R$,
-$A \le \baseof{L} \implies A \le R$, 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. $C \nothaspatch \p$.
-
-\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 some $D$ 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 $D \not\in \py, R \in \py$:}
-
-xxx up to here
-
-%D \in \py$:}