X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=b97f7bb606db5b814b129ec5561dc39ef6c62042;hb=c60a10efc12a3c6b09c536c424fc9ca24a167350;hp=b97350feb3e49447324ba7a794cdaf5b5ebe6425;hpb=d9b8ed895bd893cc9f96495b57cd0f7b8b70efa8;p=topbloke-formulae.git

diff --git a/article.tex b/article.tex
index b97350f..b97f7bb 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
@@ -248,19 +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 $C$ such that $\merge{C}{L}{M}{R}$, No Replay
-is preserved.  {\it Proof:}
+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.
 
@@ -270,9 +281,6 @@ $D \isin C$.  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 $D \le L \lor D \le
@@ -280,8 +288,7 @@ R$ 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.
+$D \not\isin C$.  OK.
 
 $\qed$
 
@@ -386,7 +393,7 @@ $D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
 \section{Anticommit}
 
 Given $L, R^+, R^-$ where
-$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.  
+$R^+ \in \pry, R^- = \baseof{R^+}$.  
 Construct $C$ which has $\pr$ removed.
 Used for removing a branch dependency.
 \gathbegin
@@ -394,7 +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}
@@ -402,16 +409,60 @@ Used for removing a branch dependency.
 \[ \eqn{ Unique Tip }{
  \pendsof{L}{\pry} = \{ R^+ \}
 }\]
-\[ \eqn{ Correct Base }{
- \baseof{R^+} = R^-
-}\]
 \[ \eqn{ Currently Included }{
  L \haspatch \pry
 }\]
+\[ \eqn{ Not Self }{
+ L \not\in \{ R^+ \}
+}\]
+
+\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.
 
-xxx want to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+$\qed$
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$.
+
+xxx tbd
+
+xxx need to finish anticommit
 
 \section{Merge}
 
@@ -421,8 +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}
 
@@ -436,10 +488,17 @@ 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{Merge Results}
+\subsection{No Replay}
 
-As above.
+See No Replay for Merge Results.
 
 \subsection{Unique Base}
 
@@ -475,4 +534,61 @@ 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
+
+Merge Ends applies.
+
+$D \isin Y \equiv D \le Y$.  $D \not\isin X$.  Recall that we
+are considering $D \in \py$.
+
+Consider $D = C$.  Thus $C \in \py, L \in \py$.  
+But $X \not\haspatch \p$ means xxx wip
+But $X \not\haspatch \p$ means $D \not\in X$, 
+
+so we have $L = Y, R =
+X$.  Thus $R \not\haspatch \p$ and by Tip Self Inpatch $R \not\in
+\py$.  Thus by Tip Merge $R \in \pn$ and $M = \baseof{L}$.
+So by Base Acyclic, $M \nothaspatch \py$.  Thus we are expecting 
+$C \haspatch \py$.  And indeed $D \isin C$ and $D \le C$.  OK.
+
 \end{document}