X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=642190f9f823431a058f3aa1b5793bc3f1751396;hp=32164de0f7c0c4ab51f14c0b99a2965ddf355fd5;hb=ae1885abbd4836b6e20ac7b7157e3cabec1d25a3;hpb=53f9c060d6ef0257657fe8aa9ddb2bb11f18d3bc

diff --git a/article.tex b/article.tex
index 32164de..642190f 100644
--- a/article.tex
+++ b/article.tex
@@ -59,10 +59,18 @@
     {\hbox{\scriptsize$\forall$}}}%
 }
 
+\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}
 
 \section{Notation}
@@ -108,7 +116,7 @@ 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} $ ]
@@ -215,12 +223,15 @@ by the LHS.  And $A \le A''$.
 \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
@@ -270,23 +281,95 @@ 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}
+\subsection{Coherence and patch inclusion}
+
+Need to consider $D \in \py$
 
 \subsubsection{For $A \haspatch P, D = C$:}
-\[ D \isin C \equiv \ldots \lor t \text{ so } D \haspatch C \]
-\[ D \le C \]
-OK
 
-\section{Test more symbols}
+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 $.
 
-$ C \haspatch \p $
+So:
+\[ A \haspatch P \implies C \haspatch P \]
 
-$ C \nothaspatch \p $
+\subsubsection{For $A \nothaspatch P$:}
 
-$ \p \patchisin C $
+Firstly, $C \not\in \py$ since if it were, $A \in \py$.  
+Thus $D \neq C$.
 
-$ \p \notpatchisin C $
+Now by contents of $A$, $D \notin A$, so $D \notin C$.
 
-$ \{ B \} \areparents 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{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
+ 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{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}
+
+\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 $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$.  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 $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \isin M$:}
+$D \not\isin C$.  Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$.  OK.
+
+$\qed$
 
 \end{document}