heading for other side

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index cae101934b5275e33c485a243e359eaca8e01e0d..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}^-}
 
@@ -391,22 +396,24 @@ 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 $A \le L \lor A \le R$.
+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 $R \in FIXME py$:}
+\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
-$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
 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 $