\documentclass[a4paper,leqno]{strayman}
+\errorcontextlines=50
\let\numberwithin=\notdef
\usepackage{amsmath}
\usepackage{mathabx}
\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}^-}
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,
$A \le L \equiv A \le \baseof{L}$.
-\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
-$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
-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 $R \in \py$:}
+$\qed$
\end{document}