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wip merge
author
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 5 Mar 2012 16:55:36 +0000
(16:55 +0000)
committer
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 5 Mar 2012 16:55:36 +0000
(16:55 +0000)
article.tex
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diff --git
a/article.tex
b/article.tex
index 4e114f280985d82f5c68d8ce1e2e639193cc560b..56198210808a9552089ec5fc7fefe7b4409d5c59 100644
(file)
--- a/
article.tex
+++ b/
article.tex
@@
-394,9
+394,9
@@
Need to consider only $C \in \py$, ie $L \in \py$,
and calculate $\pendsof{C}{\pn}$. So we will consider some
putative ancestor $A \in \pn$ and see whether $A \le C$.
and calculate $\pendsof{C}{\pn}$. So we will consider some
putative ancestor $A \in \pn$ and see whether $A \le C$.
-$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+
By Exact Ancestors for C,
$A \le C \equiv A \le L \lor A \le R \lor A = C$.
But $C \in py$ and $A \in \pn$ so $A \neq C$.
But $C \in py$ and $A \in \pn$ so $A \neq C$.
-Thus $
fixme this is not really the right thing
A \le L \lor A \le R$.
+Thus $
A \le C \equiv
A \le L \lor A \le R$.
By Unique Base of L and Transitive Ancestors,
$A \le L \equiv A \le \baseof{L}$.
By Unique Base of L and Transitive Ancestors,
$A \le L \equiv A \le \baseof{L}$.
@@
-416,7
+416,7
@@
Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} =
UP TO HERE
UP TO HERE
-By Tip Merge
,
$A \le $
+By Tip Merge
condition on
$A \le $
Let $S =
\begin{cases}
Let $S =
\begin{cases}