X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=90f5ab2ede0ee1cbc92ccaf41374242f5cb72028;hp=71ad81fbc77d71247f48bbe786242696aa6073d9;hb=3ac7463b1a5b9e66ccde2ab7f5a16a294812114e;hpb=5eb289521e72ab31220fdf9ad5256cad88f2c76f

diff --git a/merge.tex b/merge.tex
index 71ad81f..90f5ab2 100644
--- a/merge.tex
+++ b/merge.tex
@@ -111,14 +111,23 @@ $\qed$
 
 \subsection{Coherence and Patch Inclusion}
 
-Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
-This involves considering $D \in \py$.
+$$
+\begin{cases}
+  L \nothaspatch \p \land R \nothaspatch \p : & C \nothaspatch \p  \\
+  L \haspatch    \p \land R \haspatch    \p : & C \haspatch    \p  \\
+  \text{otherwise} \land M \haspatch    \p  : & C \nothaspatch \p  \\
+  \text{otherwise} \land M \nothaspatch \p  : & C \haspatch    \p
+\end{cases}
+$$
+\proofstarts
+~ Consider $D \in \py$.
 
 \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 $\neg[ L \nothaspatch \p ]$ by Tip Own Contents for $L$).
+\in \py$ ie $L \haspatch \p$ by Tip Own Contents for $L$).
 So $D \neq C$.
 Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+OK.
 
 \subsubsection{For $L \haspatch \p, R \haspatch \p$:}
 $D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
@@ -150,23 +159,18 @@ and this $F \le C$ so indeed $C \haspatch \p$.
 
 \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
 
-$M \haspatch \p \implies C \nothaspatch \p$.
-$M \nothaspatch \p \implies C \haspatch \p$.
-
-\proofstarts
-
 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
 if $M \haspatch \p$, $D \not\isin C$,
-whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
+whereas if $M \nothaspatch \p$, $D \isin C \equiv D \le C$.
 And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
 $F \le C$ so this suffices.
 
 Consider $D = C$:  Thus $C \in \py, L \in \py$.
-By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
+By Tip Own Contents, $L \haspatch \p$ so $L \neq X$,
 therefore we must have $L=Y$, $R=X$.
 By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
 by Base Acyclic $M \nothaspatch \p$.  By $\merge$, $D \isin C$,