From: Ian Jackson <ijackson@chiark.greenend.org.uk>
Date: Mon, 5 Mar 2012 19:20:05 +0000 (+0000)
Subject: No Replay for Merge Results
X-Git-Tag: f0.2~152
X-Git-Url: https://www.chiark.greenend.org.uk/ucgi/~ian/git?a=commitdiff_plain;h=fe4881a01172df0a21e3951fc1f01e73f17c50cc;p=topbloke-formulae.git

No Replay for Merge Results
---

diff --git a/article.tex b/article.tex
index f712ba6..674e593 100644
--- a/article.tex
+++ b/article.tex
@@ -254,6 +254,34 @@ by the LHS.  And $A \le A''$.
 }\]
 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:}
+
+\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.
+
+$\qed$
+
 \section{Commit annotation}
 
 We annotate each Topbloke commit $C$ with:
@@ -406,30 +434,9 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
    \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.
+\subsection{Merge Results}
 
-$\qed$
+As above.
 
 \subsection{Unique Base}