X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=d5e73b1a7e19da4ee9041dcb04a1bebcf5a7b4d4;hp=5459b88726b9e2264e74bd0261cd67bc54ae34b3;hb=7dc335c17ae313c006e2283a35ca214b213ffcd9;hpb=90658ed7a37fd3e1dd65439b54e9dc3bd4f3b503

diff --git a/article.tex b/article.tex
index 5459b88..d5e73b1 100644
--- a/article.tex
+++ b/article.tex
@@ -8,6 +8,13 @@
 \usepackage{mdwlist}
 %\usepackage{accents}
 
+\usepackage{fancyhdr}
+\pagestyle{fancy}
+\lhead[\rightmark]{}
+
+\let\stdsection\section
+\renewcommand\section{\newpage\stdsection}
+
 \renewcommand{\ge}{\geqslant}
 \renewcommand{\le}{\leqslant}
 \newcommand{\nge}{\ngeqslant}
@@ -34,6 +41,10 @@
 \newcommand{\py}{\pay{P}}
 \newcommand{\pn}{\pan{P}}
 
+\newcommand{\pq}{\pa{Q}}
+\newcommand{\pqy}{\pay{Q}}
+\newcommand{\pqn}{\pan{Q}}
+
 \newcommand{\pr}{\pa{R}}
 \newcommand{\pry}{\pay{R}}
 \newcommand{\prn}{\pan{R}}
@@ -93,605 +104,15 @@
 
 \begin{document}
 
-\section{Notation}
-
-\begin{basedescript}{
-\desclabelwidth{5em}
-\desclabelstyle{\nextlinelabel}
-}
-\item[ $ C \hasparents \set X $ ]
-The parents of commit $C$ are exactly the set
-$\set X$.
-
-\item[ $ C \ge D $ ]
-$C$ is a descendant of $D$ in the git commit
-graph.  This is a partial order, namely the transitive closure of 
-$ D \in \set X $ where $ C \hasparents \set X $.
-
-\item[ $ C \has D $ ]
-Informally, the tree at commit $C$ contains the change
-made in commit $D$.  Does not take account of deliberate reversions by
-the user or reversion, rebasing or rewinding in
-non-Topbloke-controlled branches.  For merges and Topbloke-generated
-anticommits or re-commits, the ``change made'' is only to be thought
-of as any conflict resolution.  This is not a partial order because it
-is not transitive.
-
-\item[ $ \p, \py, \pn $ ]
-A patch $\p$ consists of two sets of commits $\pn$ and $\py$, which
-are respectively the base and tip git branches.  $\p$ may be used
-where the context requires a set, in which case the statement
-is to be taken as applying to both $\py$ and $\pn$.
-None of these sets overlap.  Hence:
-
-\item[ $ \patchof{ C } $ ]
-Either $\p$ s.t. $ C \in \p $, or $\bot$.  
-A function from commits to patches' sets $\p$.
-
-\item[ $ \pancsof{C}{\set P} $ ]
-$ \{ A \; | \; A \le C \land A \in \set P \} $ 
-i.e. all the ancestors of $C$
-which are in $\set P$.
-
-\item[ $ \pendsof{C}{\set P} $ ]
-$ \{ E \; | \; E \in \pancsof{C}{\set P}
-  \land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}}
-  E \neq A \land E \le A \} $ 
-i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$.
-
-\item[ $ \baseof{C} $ ]
-$ \pendsof{C}{\pn} = \{ \baseof{C} \} $ where $ C \in \py $.
-A partial function from commits to commits.
-See Unique Base, below.
-
-\item[ $ C \haspatch \p $ ]
-$\displaystyle \bigforall_{D \in \py} D \isin C \equiv D \le C $.
-~ Informally, $C$ has the contents of $\p$.
-
-\item[ $ C \nothaspatch \p $ ]
-$\displaystyle \bigforall_{D \in \py} D \not\isin C $.
-~ Informally, $C$ has none of the contents of $\p$.  
-
-Non-Topbloke commits are $\nothaspatch \p$ for all $\p$; if a Topbloke
-patch is applied to a non-Topbloke branch and then bubbles back to
-the Topbloke patch itself, we hope that git's merge algorithm will
-DTRT or that the user will no longer care about the Topbloke patch.
-
-\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
-The contents of a git merge result:
-
-$\displaystyle D \isin C \equiv
-  \begin{cases}
-    (D \isin L \land D \isin R) \lor D = C : & \true \\
-    (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
-    \text{otherwise} : & D \not\isin M
-  \end{cases}
-$ 
-
-Some (overlapping) alternative formulations:
-
-$\displaystyle D \isin C \equiv
-  \begin{cases}
-    D \isin L \equiv D \isin R     : & D = C \lor D \isin L \\
-    D \isin L \equiv D \isin R     : & D = C \lor D \isin R \\
-    D \isin L \nequiv D \isin R  : & D = C \lor D \not\isin M \\
-    D \isin M \equiv D \isin L     : & D = C \lor D \isin R \\
-    D \isin M \equiv D \isin R     : & D = C \lor D \isin L \\
-  \end{cases}
-$
-
-\end{basedescript}
-\newpage
-\section{Invariants}
-
-We maintain these each time we construct a new commit. \\
-\[ \eqn{No Replay:}{
-  C \has D \implies C \ge D
-}\]
-\[\eqn{Unique Base:}{
- \bigforall_{C \in \py} \pendsof{C}{\pn} = \{ B \}
-}\]
-\[\eqn{Tip Contents:}{
-  \bigforall_{C \in \py} D \isin C \equiv
-    { D \isin \baseof{C} \lor \atop
-      (D \in \py \land D \le C) }
-}\]
-\[\eqn{Base Acyclic:}{
-  \bigforall_{B \in \pn} D \isin B \implies D \notin \py
-}\]
-\[\eqn{Coherence:}{
-  \bigforall_{C,\p} C \haspatch \p \lor C \nothaspatch \p
-}\]
-\[\eqn{Foreign Inclusion:}{
-  \bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
-}\]
-
-\section{Some lemmas}
-
-\[ \eqn{Exclusive Tip Contents:}{
-  \bigforall_{C \in \py} 
-    \neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C )
-      \Bigr]
-}\]
-Ie, the two limbs of the RHS of Tip Contents are mutually exclusive.
-
-\proof{
-Let $B = \baseof{C}$ in $D \isin \baseof{C}$.  Now $B \in \pn$.
-So by Base Acyclic $D \isin B \implies D \notin \py$.
-}
-\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
-  \bigforall_{C \in \py} D \isin C \equiv
-  \begin{cases}
-    D \in \py : & D \le C \\
-    D \not\in \py : & D \isin \baseof{C}
-  \end{cases}
-}\]
-
-\[ \eqn{Tip Self Inpatch:}{
-  \bigforall_{C \in \py} C \haspatch \p
-}\]
-Ie, tip commits contain their own patch.
-
-\proof{
-Apply Exclusive Tip Contents to some $D \in \py$:
-$ \bigforall_{C \in \py}\bigforall_{D \in \py}
-  D \isin C \equiv D \le C $
-}
-
-\[ \eqn{Exact Ancestors:}{
-  \bigforall_{ C \hasparents \set{R} }
-  D \le C \equiv
-    ( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
-    \lor D = C
-}\]
-
-\[ \eqn{Transitive Ancestors:}{
-  \left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv
-  \left[ \bigforall_{ A \in \pancsof{C}{\set P} } A \le M \right]
-}\]
-
-\proof{
-The implication from right to left is trivial because
-$ \pends() \subset \pancs() $.
-For the implication from left to right: 
-by the definition of $\mathcal E$,
-for every such $A$, either $A \in \pends()$ which implies
-$A \le M$ by the LHS directly,
-or $\exists_{A' \in \pancs()} \; A' \neq A \land A \le A' $
-in which case we repeat for $A'$.  Since there are finitely many
-commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
-by the LHS.  And $A \le A''$.
-}
-\[ \eqn{Calculation Of Ends:}{
-  \bigforall_{C \hasparents \set A}
-    \pendsof{C}{\set P} =
-       \left\{ E \Big|
-           \Bigl[ \Largeexists_{A \in \set A} 
-                       E \in \pendsof{A}{\set P} \Bigr] \land
-           \Bigl[ \Largenexists_{B \in \set A} 
-                       E \neq B \land E \le B \Bigr]
-       \right\}
-}\]
-XXX proof TBD.
-
-\subsection{No Replay for Merge Results}
-
-If we are constructing $C$, with,
-\gathbegin
-  \mergeof{C}{L}{M}{R}
-\gathnext
-  L \le C
-\gathnext
-  R \le C
-\end{gather}
-No Replay is preserved.  \proofstarts
-
-\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 $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \not\isin M$:}
-$D \isin C$.  Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
-R$ so $D \le C$.  OK.
-
-\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \isin M$:}
-$D \not\isin C$.  OK.
-
-$\qed$
-
-\section{Commit annotation}
-
-We annotate each Topbloke commit $C$ with:
-\gathbegin
- \patchof{C}
-\gathnext
- \baseof{C}, \text{ if } C \in \py
-\gathnext
- \bigforall_{\pa{Q}} 
-   \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
-\gathnext
- \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
-\end{gather}
-
-We do not annotate $\pendsof{C}{\py}$ for $C \in \py$.  Doing so would
-make it wrong to make plain commits with git because the recorded $\pends$
-would have to be updated.  The annotation is not needed because
-$\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
-
-\section{Simple commit}
-
-A simple single-parent forward commit $C$ as made by git-commit.
-\begin{gather}
-\tag*{} C \hasparents \{ A \} \\
-\tag*{} \patchof{C} = \patchof{A} \\
-\tag*{} D \isin C \equiv D \isin A \lor D = C
-\end{gather}
-
-\subsection{No Replay}
-Trivial.
-
-\subsection{Unique Base}
-If $A, C \in \py$ then $\baseof{C} = \baseof{A}$. $\qed$
-
-\subsection{Tip Contents}
-We need to consider only $A, C \in \py$.  From Tip Contents for $A$:
-\[ D \isin A \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \]
-Substitute into the contents of $C$:
-\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A )
-    \lor D = C \]
-Since $D = C \implies D \in \py$, 
-and substituting in $\baseof{C}$, this gives:
-\[ D \isin C \equiv D \isin \baseof{C} \lor
-    (D \in \py \land D \le A) \lor
-    (D = C \land D \in \py) \]
-\[ \equiv D \isin \baseof{C} \lor
-   [ D \in \py \land ( D \le A \lor D = C ) ] \]
-So by Exact Ancestors:
-\[ D \isin C \equiv D \isin \baseof{C} \lor ( D \in \py \land D \le C
-) \]
-$\qed$
-
-\subsection{Base Acyclic}
-
-Need to consider only $A, C \in \pn$.  
-
-For $D = C$: $D \in \pn$ so $D \not\in \py$. OK.
-
-For $D \neq C$: $D \isin C \equiv D \isin A$, so by Base Acyclic for
-$A$, $D \isin C \implies D \not\in \py$. $\qed$
-
-\subsection{Coherence and patch inclusion}
-
-Need to consider $D \in \py$
-
-\subsubsection{For $A \haspatch P, D = C$:}
-
-Ancestors of $C$:
-$ D \le C $.
-
-Contents of $C$:
-$ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $.
-
-\subsubsection{For $A \haspatch P, D \neq C$:}
-Ancestors: $ D \le C \equiv D \le A $.
-
-Contents: $ D \isin C \equiv D \isin A \lor f $
-so $ D \isin C \equiv D \isin A $.
-
-So:
-\[ A \haspatch P \implies C \haspatch P \]
-
-\subsubsection{For $A \nothaspatch P$:}
-
-Firstly, $C \not\in \py$ since if it were, $A \in \py$.  
-Thus $D \neq C$.
-
-Now by contents of $A$, $D \notin A$, so $D \notin C$.
-
-So:
-\[ A \nothaspatch P \implies C \nothaspatch P \]
-$\qed$
-
-\subsection{Foreign inclusion:}
-
-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}
-
-\[ \eqn{ Tip Merge }{
- L \in \py \implies
-   \begin{cases}
-      R \in \py : & \baseof{R} \ge \baseof{L}
-              \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
-      R \in \pn : & M = \baseof{L} \\
-      \text{otherwise} : & \false
-   \end{cases}
-}\]
-\[ \eqn{ Merge Acyclic }{
-    L \in \pn
-   \implies
-    R \nothaspatch \p
-}\]
-\[ \eqn{ Removal Merge Ends }{
-    X \not\haspatch \p \land
-    Y \haspatch \p \land
-    M \haspatch \p
-  \implies
-    \pendsof{Y}{\py} = \pendsof{M}{\py}
-}\]
-\[ \eqn{ Addition Merge Ends }{
-    X \not\haspatch \p \land
-    Y \haspatch \p \land
-    M \nothaspatch \p
-   \implies \left[
-    \bigforall_{E \in \pendsof{X}{\py}} E \le Y
-   \right]
-}\]
-
-\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,
-and coherence and patch inclusion, of $C$ 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$:}
-
-
-
-xxx the coherence is not that useful ?
-
-$L \haspatch \p$ by 
-
-xxx need to recheck this
-
-$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$.  OK.
-
-\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
-
-Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
-
-Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
+\input{notation.tex}
+\input{invariants.tex}
+\input{lemmas.tex}
+\input{annotations.tex}
 
-But by Tip Merge, $\baseof{R} \ge \baseof{L}$
+\input{simple.tex}
+\input{create-base.tex}
+\input{create-tip.tex}
+\input{anticommit.tex}
+\input{merge.tex}
 
 \end{document}