X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=merge.tex;h=94b38fe9648d88f2022ba9ee57aae96d35806a2e;hb=45d3a192063cc2e12dea5fc84a8ffcf7c2a220e2;hp=fc5df3bbba880e392eccee6b0da74581b7f04a42;hpb=bc5b48777942f1a3504812ef1822febf4c354984;p=topbloke-formulae.git

diff --git a/merge.tex b/merge.tex
index fc5df3b..94b38fe 100644
--- a/merge.tex
+++ b/merge.tex
@@ -111,20 +111,29 @@ $\qed$
 
 \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 $D \isin R \equiv D \le R$.
 (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 $C \zhaspatch \p$.
 
 For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
  \equiv D \isin L \lor D \isin R$.
@@ -132,27 +141,24 @@ 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 $C \zhaspatch \p$.
 
 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 $C \zhaspatch \p$.
 
 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 $C \zhaspatch \p$.
 
-So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+So, in all cases, $C \zhaspatch \p$.
+And by $L \haspatch \p$, $\exists_{F \in \py} F \le L$
+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 $D \in \py$.
 $D \isin Y \equiv D \le Y$.  $D \not\isin X$.
@@ -160,6 +166,8 @@ We will show for each of
 various cases that
 if $M \haspatch \p$, $D \not\isin C$,
 whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
+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 $L \neq X$,