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more clarifications and fixes from reread
author
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 12 Mar 2012 17:59:28 +0000
(17:59 +0000)
committer
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 12 Mar 2012 17:59:28 +0000
(17:59 +0000)
article.tex
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a/article.tex
b/article.tex
index 40c65188362c926e492f2b458331f1714e3b8069..321914746537919ca730ce384393fe522d268862 100644
(file)
--- a/
article.tex
+++ b/
article.tex
@@
-640,7
+640,7
@@
$$
\subsubsection{For $\p = \pq$:}
By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
-By No Sneak, $D \
le B
\equiv D = C$. Thus $C \haspatch \pq$.
+By No Sneak, $D \
not\le B$ so $D \le C
\equiv D = C$. Thus $C \haspatch \pq$.
\subsubsection{For $\p \neq \pq$:}