wip merge complex - a fix

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index a0adacbdee9b8d02bb1f2421f8158b1a43b505b2..b97f7bb606db5b814b129ec5561dc39ef6c62042 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -488,6 +488,13 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
       \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{No Replay}
 
@@ -529,19 +536,19 @@ $\qed$
 
 \subsection{Coherence and patch inclusion}
 
-Need to determine $C \haspatch P$ based on $L,M,R \haspatch P$.
+Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
 This involves considering $D \in \py$.  
 
-\subsubsection{For $L \nothaspatch P, R \nothaspatch P$:}
+\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. $C \nothaspatch P$.
+\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 $L \haspatch P, R \haspatch P$:}
+\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 $C \haspatch P$.
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
 
 For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
  \equiv D \isin L \lor D \isin R$.  
@@ -549,18 +556,39 @@ For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
 
 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 $C \haspatch P$.
+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 $C \haspatch P$.
+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 $C \haspatch P$.
+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 indeed $L \haspatch P \land R \haspatch P \implies C \haspatch P$.
+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}