wip merge complex - reformulate merge ends conditions

diff --git a/article.tex b/article.tex
index 1feeb4834899e942a4d300f7d48df120e58426ec..b71b1884bacd8291b3e4c8016453e55b6e910adb 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -488,16 +488,20 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
       \text{otherwise} : & \false
    \end{cases}
 }\]
-\[ \eqn{ Merge Ends }{
+\[ \eqn{ Removal Merge Ends }{
     X \not\haspatch \p \land
-    Y \haspatch \p
+    Y \haspatch \p \land
+    M \haspatch \p
   \implies
-  \begin{cases}
-    M \haspatch \p :    & \pendsof{Y}{\py} = \pendsof{M}{\py}
-   \\
-    M \nothaspatch \p : & \displaystyle
-                          \bigforall_{E \in \pendsof{X}{\py}} E \le Y
-  \end{cases}
+    \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+    X \not\haspatch \p \land
+    Y \haspatch \p \land
+    M \nothaspatch \p
+   \implies \left[
+    \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+   \right]
 }\]
 
 \subsection{No Replay}
@@ -580,7 +584,8 @@ $C \haspatch \p \equiv M \nothaspatch \p$.
 
 \proofstarts
 
-Merge Ends applies.  Recall that we are considering $D \in \py$.
+One of the Merge Ends conditions applies.  
+Recall that we are considering $D \in \py$.
 $D \isin Y \equiv D \le Y$.  $D \not\isin X$.
 We will show for each of
 various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
@@ -597,12 +602,12 @@ $D \not\isin M$ so by $\merge$, $D \isin C$.  OK.
 
 Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
 $D \not\le Y$.  If $D \le X$ then
-$D \in \pancsof{X}{\py}$, so by Merge Ends and 
+$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and 
 Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
 Thus $D \not\le C$.  By $\merge$, $D \not\isin C$.  OK.
 
 Consider $D \neq C, M \haspatch P, D \isin Y$:
-$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Merge Ends
+$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
 and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
 Thus $D \isin M$.  By $\merge$, $D \not\isin C$.  OK.