introduce \proofstarts

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 023a18a9340c7670e3ee89011177eb9010165e69..1fa67c2330d692aed86011628931177731e8a99a 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -81,7 +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*{}}
@@ -270,7 +271,7 @@ If we are constructing $C$, given
 \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.
 
@@ -424,7 +425,7 @@ Merge Results applies. $\qed$
 \subsection{Desired Contents}
 
 \[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \]
-{\it Proof.}
+\proofstarts
 
 \subsubsection{For $D = C$:}
 
@@ -527,8 +528,40 @@ $\qed$
 
 \subsection{Coherence and patch inclusion}
 
-xxx tbd
+Need to determine $C \haspatch P$ based on $L,M,R \haspatch P$.
+This involves considering $D \in \py$.  
+
+We will use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
+
+\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 $C \haspatch P$.
 
-xxx need to finish merge
+So indeed $L \haspatch P \land R \haspatch P \implies C \haspatch P$.
 
 \end{document}