fix title etc. in merge use of no replay

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 06b76126865f6c828a0fb58a3f7cbf9ae6d6c5b0..89d23fe179d9bfc53524f3cfe65fb20d4425f51e 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -56,8 +56,11 @@
 \newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
 %\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
 
-\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
-\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
+\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.}} }
@@ -254,6 +257,37 @@ by the LHS.  And $A \le A''$.
 }\]
 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:
@@ -355,7 +389,7 @@ $D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
 \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
@@ -363,17 +397,33 @@ Used for removing a branch dependency.
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- D \isin C \equiv
-   \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}
+ \merge{C}{L}{R^+}{R^-}
 \end{gather}
 
-xxx want to prove $D \isin C \equiv D \not\in pry \land D \isin L$.
+\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}
 
@@ -383,12 +433,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 \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}
+ \merge{C}{L}{M}{R}
 \end{gather}
 
 \subsection{Conditions}
@@ -406,28 +451,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 
 \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}