From: Ian Jackson <ijackson@chiark.greenend.org.uk>
Date: Tue, 6 Mar 2012 17:46:30 +0000 (+0000)
Subject: use \p rather than just P
X-Git-Tag: f0.2~136
X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=commitdiff_plain;h=e600e19a18c0f5b4595c68dd305f093b62fed9e4

use \p rather than just P
---

diff --git a/article.tex b/article.tex
index 2f512f4..6a5023c 100644
--- a/article.tex
+++ b/article.tex
@@ -536,19 +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 $D \neq C$: $D \le C \equiv D \le L \lor D \le R
  \equiv D \isin L \lor D \isin R$.  
@@ -556,23 +556,23 @@ 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 $L \haspatch P \land R \haspatch P \implies C \haspatch P$.
+So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
 
-\subsubsection{For (wlog) $X \not\haspatch P, Y \haspatch P$:}
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
 
-$C \haspatch P \equiv C \nothaspatch M$.
+$C \haspatch \p \equiv C \nothaspatch M$.
 
 \proofstarts