From e284656ebfa338be8bc3a02de60391063d9f8f3d Mon Sep 17 00:00:00 2001
From: Ian Jackson <ijackson@chiark.greenend.org.uk>
Date: Mon, 28 May 2012 00:20:12 +0100
Subject: [PATCH] psueomerge: sort out foreign ends

---
 pseudomerge.tex | 12 ++----------
 1 file changed, 2 insertions(+), 10 deletions(-)

diff --git a/pseudomerge.tex b/pseudomerge.tex
index 0decb5f..0f93abc 100644
--- a/pseudomerge.tex
+++ b/pseudomerge.tex
@@ -25,23 +25,15 @@ but whose contents are exactly those of $L$.
 }\]
 
 \[ \eqn{ Foreign Unaffected }{
- \bigforall_{ D \in \foreign }
-  \left[ \bigexists_{A \in \set A} D \le A \right]
-   \implies
-  D \le L
+ \pendsof{C}{\foreign} = \pendsof{L}{\foreign}
 }\]
- TODO THAT IS IMPOSSIBLE TO CALCULATE
 
 \subsection{Lemma: Foreign Identical}
 
 $\isforeign{D} \implies \big[ D \le C \equiv D \le L \big]$.
 
 \proof{
-If $D \le L$, trivially $D \le C$; so conversely
-$D \not\le C \implies D \not\le L$.  
-Whereas if $D \le C$, either $D \le L$ or
-$\exists{A \in \set A} D \le A$ (since $D \neq C$),
-in which case by Foreign Unaffected $D \le L$.
+Trivial by Foreign Unaffected and the definition of $\pends$
 }
 
 \subsection{No Replay}
-- 
2.30.2