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author
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Sun, 4 Mar 2012 20:55:43 +0000
(20:55 +0000)
committer
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Sun, 4 Mar 2012 20:55:43 +0000
(20:55 +0000)
article.tex
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a/article.tex
b/article.tex
index c705f61c638523e676692f6abd367e51d2e82cd4..4e114f280985d82f5c68d8ce1e2e639193cc560b 100644
(file)
--- a/
article.tex
+++ b/
article.tex
@@
-412,6
+412,8
@@
$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} =
\baseof{R}$.
+\subsubsection{For $R \in \pn$:}
+
UP TO HERE
By Tip Merge, $A \le $