finish merge unique base (we think)

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index f0bbf33b91975bcb61866b8eeef73fabcea17af2..1e42670e628e92e064dd162a960f8ba65b5c8c81 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}^-}
 
@@ -389,9 +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 $A \le C$.
 
-$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+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 $A \neq C$.  
-Thus $A \le L \lor A \le R$.
+Thus $A \le C \equiv A \le L \lor A \le R$.
 
 By Unique Base of L and Transitive Ancestors,
 $A \le L \equiv A \le \baseof{L}$.
@@ -403,33 +408,18 @@ $A \le R \equiv A \le \baseof{R}$.
 
 But by Tip Merge condition on $\baseof{R}$,
 $A \le \baseof{L} \implies A \le \baseof{R}$, so
-$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$.  Ie, $\baseof{C} =
-\baseof{R}$.
-
-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 
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.  
+That is, $\baseof{C} = \baseof{R}$.
 
-Let $A \in \pends{C}{\pn}$.  
-Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
-\pendsof{R,\pn}$.
+\subsubsection{For $R \in \pn$:}
 
+By Tip Merge condition on $R$,
+$A \le \baseof{L} \implies A \le R$, so
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.  
+Thus $A \le C \equiv A \le R$.  
+That is, $\baseof{C} = R$.
 
-
-%$\pends{C,
-
-%%\subsubsection{For $R \in \py$:}
-%foo
+$\qed$
 
 \end{document}