heading for other side

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 9e58393b58e0edfbdd414cfe3b4428fea2fd9325..4e114f280985d82f5c68d8ce1e2e639193cc560b 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -1,4 +1,5 @@
 \documentclass[a4paper,leqno]{strayman}
+\errorcontextlines=50
 \let\numberwithin=\notdef
 \usepackage{amsmath}
 \usepackage{mathabx}
@@ -18,8 +19,12 @@
 \newcommand{\haspatch}{\sqSupset}
 \newcommand{\patchisin}{\sqSubset}
 
-\newcommand{\set}[1]{\mathbb #1}
-\newcommand{\pa}[1]{\varmathbb #1}
+        \newif\ifhidehack\hidehackfalse
+        \DeclareRobustCommand\hidefromedef[2]{%
+          \hidehacktrue\ifhidehack#1\else#2\fi\hidehackfalse}
+        \newcommand{\pa}[1]{\hidefromedef{\varmathbb{#1}}{#1}}
+
+\newcommand{\set}[1]{\mathbb{#1}}
 \newcommand{\pay}[1]{\pa{#1}^+}
 \newcommand{\pan}[1]{\pa{#1}^-}
 
@@ -386,9 +391,51 @@ $\qed$
 \subsection{Unique Base}
 
 Need to consider only $C \in \py$, ie $L \in \py$,
-and calculate $\pendsof{C}{\pn}$.
+and calculate $\pendsof{C}{\pn}$.  So we will consider some
+putative ancestor $A \in \pn$ and see whether $A \le C$.
+
+$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+But $C \in py$ and $A \in \pn$ so $A \neq C$.  
+Thus $fixme this is not really the right thing A \le L \lor A \le R$.
+
+By Unique Base of L and Transitive Ancestors,
+$A \le L \equiv A \le \baseof{L}$.
+
+\subsubsection{For $R \in \py$:}
+
+By Unique Base of $R$ and Transitive Ancestors,
+$A \le R \equiv A \le \baseof{R}$.
+
+But by Tip Merge condition on $\baseof{R}$,
+$A \le \baseof{L} \implies A \le \baseof{R}$, so
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+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 $
+
+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 $R \in \py$:}
-%foo
+%%\subsubsection{For $R \in \py$:}
 
 \end{document}