X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=89d23fe179d9bfc53524f3cfe65fb20d4425f51e;hp=4cb6c96b79aa3fb9c5b7039feb222305a8e06ce6;hb=9e53deabff04c90177d8282bdf7969545e6dc5ee;hpb=b49f4a4167ff535af48ccaba87e7ff8a02b77347

diff --git a/article.tex b/article.tex
index 4cb6c96..89d23fe 100644
--- a/article.tex
+++ b/article.tex
@@ -1,4 +1,5 @@
 \documentclass[a4paper,leqno]{strayman}
+\errorcontextlines=50
 \let\numberwithin=\notdef
 \usepackage{amsmath}
 \usepackage{mathabx}
@@ -18,8 +19,12 @@
 \newcommand{\haspatch}{\sqSupset}
 \newcommand{\patchisin}{\sqSubset}
 
-\newcommand{\set}[1]{\mathbb #1}
-\newcommand{\pa}[1]{\varmathbb #1}
+        \newif\ifhidehack\hidehackfalse
+        \DeclareRobustCommand\hidefromedef[2]{%
+          \hidehacktrue\ifhidehack#1\else#2\fi\hidehackfalse}
+        \newcommand{\pa}[1]{\hidefromedef{\varmathbb{#1}}{#1}}
+
+\newcommand{\set}[1]{\mathbb{#1}}
 \newcommand{\pay}[1]{\pa{#1}^+}
 \newcommand{\pan}[1]{\pa{#1}^-}
 
@@ -27,6 +32,10 @@
 \newcommand{\py}{\pay{P}}
 \newcommand{\pn}{\pan{P}}
 
+\newcommand{\pr}{\pa{R}}
+\newcommand{\pry}{\pay{R}}
+\newcommand{\prn}{\pan{R}}
+
 %\newcommand{\hasparents}{\underaccent{1}{>}}
 %\newcommand{\hasparents}{{%
 %  \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}}
@@ -44,8 +53,14 @@
 \newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
 \newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
 
-\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
-\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
+\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
+%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
+
+\newcommand{\patch}{{\mathcal P}}
+\newcommand{\base}{{\mathcal B}}
+
+\newcommand{\patchof}[1]{\patch ( #1 ) }
+\newcommand{\baseof}[1]{\base ( #1 ) }
 
 \newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
 \newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
@@ -59,7 +74,17 @@
     {\hbox{\scriptsize$\forall$}}}%
 }
 
-\newcommand{\proof}[1]{{\it Proof.} #1 $\square$}
+\newcommand{\Largeexists}{\mathop{\hbox{\Large$\exists$}}}
+\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
+
+\newcommand{\qed}{\square}
+\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+
+\newcommand{\gathbegin}{\begin{gather} \tag*{}}
+\newcommand{\gathnext}{\\ \tag*{}}
+
+\newcommand{\true}{t}
+\newcommand{\false}{f}
 
 \begin{document}
 
@@ -106,13 +131,13 @@ 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}}
-  A \neq E \land E \le A \} $ 
+  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.
+See Unique Base, below.
 
 \item[ $ C \haspatch \p $ ]
 $\displaystyle \bigforall_{D \in \py} D \isin C \equiv D \le C $.
@@ -127,6 +152,17 @@ 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 \merge{C}{L}{M}{R} $ ]
+The contents of a git merge result:
+
+$\displaystyle D \isin C \equiv
+  \begin{cases}
+    (D \isin L \land D \isin R) \lor D = C : & \true \\
+    (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
+    \text{otherwise} : & D \not\isin M
+  \end{cases}
+$ 
+
 \end{basedescript}
 \newpage
 \section{Invariants}
@@ -209,16 +245,61 @@ 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 $A \le A''$.
 }
+\[ \eqn{Calculation Of Ends:}{
+  \bigforall_{C \hasparents \set A}
+    \pendsof{C}{\set P} =
+       \Bigl\{ E \Big|
+           \Bigl[ \Largeexists_{A \in \set A} 
+                       E \in \pendsof{A}{\set P} \Bigr] \land
+           \Bigl[ \Largenexists_{B \in \set A} 
+                       E \neq B \land E \le B \Bigr]
+       \Bigr\}
+}\]
+XXX proof TBD.
+
+\subsection{No Replay for Merge Results}
+
+If we are constructing $C$, given
+\gathbegin
+  \merge{C}{L}{M}{R}
+\gathnext
+  L \le C
+\gathnext
+  R \le C
+\end{gather}
+No Replay is preserved.  {\it Proof:}
+
+\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:
-\begin{gather}
-\tag*{} \patchof{C} \\
-\tag*{} \baseof{C}, \text{ if } C \in \py \\
-\tag*{} \bigforall_{\pa{Q}} 
-        \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q} \\
-\tag*{} \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+\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
@@ -226,16 +307,184 @@ 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{Test more symbols}
+\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
+ \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{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 $R \in \py$:}
 
-$ C \haspatch \p $
+By Unique Base of $R$ and Transitive Ancestors,
+$A \le R \equiv A \le \baseof{R}$.
 
-$ C \nothaspatch \p $
+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, $\baseof{C} = \baseof{R}$.
 
-$ \p \patchisin C $
+\subsubsection{For $R \in \pn$:}
 
-$ \p \notpatchisin C $
+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, $\baseof{C} = R$.
 
-$ \{ B \} \areparents C $
+$\qed$
 
 \end{document}