formatting fix

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 56198210808a9552089ec5fc7fefe7b4409d5c59..4c5f6751682c83a13e478fdb11d64496abd1bd22 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -32,6 +32,10 @@
 \newcommand{\py}{\pay{P}}
 \newcommand{\pn}{\pan{P}}
 
 \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{>}}}
 %\newcommand{\hasparents}{\underaccent{1}{>}}
 %\newcommand{\hasparents}{{%
 %  \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}}


@@ -49,8 +53,14 @@
 \newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
 \newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
 
 \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.}} }
 
 \newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
 \newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }


@@ -142,6 +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.
 
 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}
 \end{basedescript}
 \newpage
 \section{Invariants}


@@ -227,15 +248,46 @@ by the LHS.  And $A \le A''$.
 \[ \eqn{Calculation Of Ends:}{
   \bigforall_{C \hasparents \set A}
     \pendsof{C}{\set P} =
 \[ \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]
            \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.
 
 }\]
 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:
 \section{Commit annotation}
 
 We annotate each Topbloke commit $C$ with:


@@ -334,6 +386,45 @@ $\qed$
 If $D = C$, trivial.  For $D \neq C$:
 $D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
 
 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 $M$ ($M < L, M < R$):
 \section{Merge}
 
 Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):


@@ -342,12 +433,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
 \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}
+ \merge{C}{L}{M}{R}

 \end{gather}
 
 \subsection{Conditions}
 \end{gather}
 
 \subsection{Conditions}


@@ -365,28 +451,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 
 \subsection{No Replay}
 
 
 \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.
-
-$\qed$
+See No Replay for Merge Results.

 
 \subsection{Unique Base}
 
 
 \subsection{Unique Base}
 


@@ -409,33 +474,17 @@ $A \le R \equiv A \le \baseof{R}$.
 But by Tip Merge condition on $\baseof{R}$,
 $A \le \baseof{L} \implies A \le \baseof{R}$, so
 $A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
 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 $A \le C \equiv A \le \baseof{R}$.  Ie, $\baseof{C} =
-\baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.  
+That is, $\baseof{C} = \baseof{R}$.

 
 \subsubsection{For $R \in \pn$:}
 
 
 \subsubsection{For $R \in \pn$:}
 

-UP TO HERE
-
-By Tip Merge condition on $A \le $
-
-Let $S =
-   \begin{cases} 
-     R \in \py : & \baseof{R} \\
-     R \in \pn : & R
-   \end{cases}$.  
-Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
-   
-Consider some $A \in \pn$.  If $A \le S$ then $A \le C$.
-If $A \not\le S$ then 
+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, $\baseof{C} = R$.

 
 

-Let $A \in \pends{C}{\pn}$.  
-Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
-\pendsof{R,\pn}$.
-
-
-
-%$\pends{C,
-
-%%\subsubsection{For $R \in \py$:}
+$\qed$

 
 \end{document}
 
 \end{document}