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finish merge unique base (we think)
author
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 5 Mar 2012 18:07:15 +0000
(18:07 +0000)
committer
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Mon, 5 Mar 2012 18:07:15 +0000
(18:07 +0000)
article.tex
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diff --git
a/article.tex
b/article.tex
index f8acb2260ffcf7898dd228718269b1efe9f40447..1e42670e628e92e064dd162a960f8ba65b5c8c81 100644
(file)
--- a/
article.tex
+++ b/
article.tex
@@
-409,34
+409,17
@@
$A \le R \equiv A \le \baseof{R}$.
But by Tip Merge condition on $\baseof{R}$,
$A \le \baseof{L} \implies A \le \baseof{R}$, so
$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
But by Tip Merge condition on $\baseof{R}$,
$A \le \baseof{L} \implies A \le \baseof{R}$, so
$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$.
Ie, $\baseof{C} =
-\baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.
+
That is, $\baseof{C} =
\baseof{R}$.
\subsubsection{For $R \in \pn$:}
By Tip Merge condition on $R$,
\subsubsection{For $R \in \pn$:}
By Tip Merge condition on $R$,
-$A \le \baseof{L} \implies A \le R$
+$A \le \baseof{L} \implies A \le R$, so
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.
+Thus $A \le C \equiv A \le R$.
+That is, $\baseof{C} = R$.
-UP TO HERE
-
-Let $S =
- \begin{cases}
- R \in \py : & \baseof{R} \\
- R \in \pn : & R
- \end{cases}$.
-Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
-
-Consider some $A \in \pn$. If $A \le S$ then $A \le C$.
-If $A \not\le S$ then
-
-Let $A \in \pends{C}{\pn}$.
-Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
-\pendsof{R,\pn}$.
-
-
-
-%$\pends{C,
-
-%%\subsubsection{For $R \in \py$:}
+$\qed$
\end{document}
\end{document}