X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=6a5023c3ceaf24813e16dfbe0d309a2a38e22178;hp=89d23fe179d9bfc53524f3cfe65fb20d4425f51e;hb=e600e19a18c0f5b4595c68dd305f093b62fed9e4;hpb=9e53deabff04c90177d8282bdf7969545e6dc5ee

diff --git a/article.tex b/article.tex
index 89d23fe..6a5023c 100644
--- a/article.tex
+++ b/article.tex
@@ -10,6 +10,8 @@
 
 \renewcommand{\ge}{\geqslant}
 \renewcommand{\le}{\leqslant}
+\newcommand{\nge}{\ngeqslant}
+\newcommand{\nle}{\nleqslant}
 
 \newcommand{\has}{\sqsupseteq}
 \newcommand{\isin}{\sqsubseteq}
@@ -53,7 +55,8 @@
 \newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
 \newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
 
-\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
+\newcommand{\merge}{{\mathcal M}}
+\newcommand{\mergeof}[4]{\merge(#1,#2,#3,#4)}
 %\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
 
 \newcommand{\patch}{{\mathcal P}}
@@ -78,7 +81,8 @@
 \newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
 
 \newcommand{\qed}{\square}
-\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+\newcommand{\proofstarts}{{\it Proof:}}
+\newcommand{\proof}[1]{\proofstarts #1 $\qed$}
 
 \newcommand{\gathbegin}{\begin{gather} \tag*{}}
 \newcommand{\gathnext}{\\ \tag*{}}
@@ -152,7 +156,7 @@ 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} $ ]
+\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
 The contents of a git merge result:
 
 $\displaystyle D \isin C \equiv
@@ -248,26 +252,26 @@ by the LHS.  And $A \le A''$.
 \[ \eqn{Calculation Of Ends:}{
   \bigforall_{C \hasparents \set A}
     \pendsof{C}{\set P} =
-       \Bigl\{ E \Big|
+       \left\{ E \Big|
            \Bigl[ \Largeexists_{A \in \set A} 
                        E \in \pendsof{A}{\set P} \Bigr] \land
            \Bigl[ \Largenexists_{B \in \set A} 
                        E \neq B \land E \le B \Bigr]
-       \Bigr\}
+       \right\}
 }\]
 XXX proof TBD.
 
 \subsection{No Replay for Merge Results}
 
-If we are constructing $C$, given
+If we are constructing $C$, with,
 \gathbegin
-  \merge{C}{L}{M}{R}
+  \mergeof{C}{L}{M}{R}
 \gathnext
   L \le C
 \gathnext
   R \le C
 \end{gather}
-No Replay is preserved.  {\it Proof:}
+No Replay is preserved.  \proofstarts
 
 \subsubsection{For $D=C$:} $D \isin C, D \le C$.  OK.
 
@@ -397,7 +401,7 @@ Used for removing a branch dependency.
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- \merge{C}{L}{R^+}{R^-}
+ \mergeof{C}{L}{R^+}{R^-}
 \end{gather}
 
 \subsection{Conditions}
@@ -408,10 +412,9 @@ Used for removing a branch dependency.
 \[ \eqn{ Currently Included }{
  L \haspatch \pry
 }\]
-
-\subsection{Desired Contents}
-
-xxx need to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+\[ \eqn{ Not Self }{
+ L \not\in \{ R^+ \}
+}\]
 
 \subsection{No Replay}
 
@@ -419,12 +422,48 @@ 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 $M$ ($M < L, M < R$):
@@ -433,8 +472,9 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- \merge{C}{L}{M}{R}
+ \mergeof{C}{L}{M}{R}
 \end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
 
 \subsection{Conditions}
 
@@ -448,6 +488,13 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
       \text{otherwise} : & \false
    \end{cases}
 }\]
+\[ \eqn{ Merge Ends }{
+    X \not\haspatch \p \land
+    Y \haspatch \p \land
+    E \in \pendsof{X}{\py}
+  \implies
+    E \le Y
+}\]
 
 \subsection{No Replay}
 
@@ -487,4 +534,46 @@ 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) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$C \haspatch \p \equiv C \nothaspatch M$.
+
+\proofstarts
+
 \end{document}