wip merge complex - a bug

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 714578dad5eba4bd461cb2db12c3c74abe3b3327..b2f943b04c0b051059c70ab61d44a8cb8535d276 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -490,10 +490,11 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
 }\]
 \[ \eqn{ Merge Ends }{
     X \not\haspatch \p \land
-    Y \haspatch \p \land
-    E \in \pendsof{X}{\py}
-  \implies
+    Y \haspatch \p
+  \implies \left[
+  \bigforall_{E \in \pendsof{X}{\py}}
     E \le Y
+  \right]
 }\]
 
 \subsection{No Replay}
@@ -572,14 +573,48 @@ So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
 
 \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
 
-$C \haspatch \p \equiv C \nothaspatch M$.
+$C \haspatch \p \equiv M \nothaspatch \p$.
 
 \proofstarts
 
-Merge Ends applies.
-
+Merge Ends applies.  Recall that we are considering $D \in \py$.
 $D \isin Y \equiv D \le Y$.  $D \not\isin X$.
-
-Consider $D = C$.  
+We will show for each of
+various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
+(which suffices by definition of $\haspatch$ and $\nothaspatch$).
+
+Consider $D = C$:  Thus $C \in \py, L \in \py$, and by Tip
+Self Inpatch $L \haspatch \p$, so $L=Y, R=X$.  By Tip Merge,
+$M=\baseof{L}$.  So by Base Acyclic $D \not\isin M$, i.e.
+$M \nothaspatch \p$.  And indeed $D \isin C$ and $D \le C$.  OK.
+
+Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+$D \le Y$ so $D \le C$.  
+$D \not\isin M$ 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 
+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 $D \le M$
+
+%anyway     D \not\isin X
+%           D \isin Y
+%           D \le Y
+
+%bad case
+%           D \not\le M
+%           D \not\isin M
+% results
+%           D \isin C    wrong
+
+%ok case
+%           D \le M
+%           D \isin M
+% results
+%           D \not\isin C  OK
 
 \end{document}