From 90658ed7a37fd3e1dd65439b54e9dc3bd4f3b503 Mon Sep 17 00:00:00 2001
From: Ian Jackson <ijackson@chiark.greenend.org.uk>
Date: Sun, 11 Mar 2012 10:32:28 +0000
Subject: [PATCH] merge remove redundant element in condition

---
 article.tex | 5 ++---
 1 file changed, 2 insertions(+), 3 deletions(-)

diff --git a/article.tex b/article.tex
index 3da2db9..5459b88 100644
--- a/article.tex
+++ b/article.tex
@@ -496,8 +496,7 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
    \begin{cases}
       R \in \py : & \baseof{R} \ge \baseof{L}
               \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
-      R \in \pn : & R \ge \baseof{L}
-              \land M = \baseof{L} \\
+      R \in \pn : & M = \baseof{L} \\
       \text{otherwise} : & \false
    \end{cases}
 }\]
@@ -552,7 +551,7 @@ That is, $\baseof{C} = \baseof{R}$.
 
 \subsubsection{For $R \in \pn$:}
 
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
 $A \le \baseof{L} \implies A \le R$, so
 $A \le R \lor A \le \baseof{L} \equiv A \le R$.  
 Thus $A \le C \equiv A \le R$.  
-- 
2.30.2