\newcommand{\patchof}[1]{\patch ( #1 ) }
\newcommand{\baseof}[1]{\base ( #1 ) }
+\newcommand{\eqntag}[2]{ #2 \tag*{\mbox{#1}} }
\newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
-\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
%\newcommand{\bigforall}{\mathop{\hbox{\huge$\forall$}}}
\newcommand{\bigforall}{%
are respectively the base and tip git branches. $\p$ may be used
where the context requires a set, in which case the statement
is to be taken as applying to both $\py$ and $\pn$.
-All these sets are distinct. Hence:
+None of these sets overlap. Hence:
\item[ $ \patchof{ C } $ ]
Either $\p$ s.t. $ C \in \p $, or $\bot$.
Let $B = \baseof{C}$ in $D \isin \baseof{C}$. Now $B \in \pn$.
So by Base Acyclic $D \isin B \implies D \notin \py$.
}
-\[ \corrolary{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
\bigforall_{C \in \py} D \isin C \equiv
\begin{cases}
D \in \py : & D \le C \\
\text{otherwise} : & \false
\end{cases}
}\]
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
\[ \eqn{ Removal Merge Ends }{
X \not\haspatch \p \land
Y \haspatch \p \land
$\qed$
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
+R$. And $D \neq C$. So $D \not\isin C$. $\qed$
+
+\subsection{Tip Contents}
+
+We will consider some $D$ and prove the Exclusive Tip Contents form.
+We use the Coherence of $C$ as just proved.
+
+xxx the coherence is not that useful ?
+
+\subsubsection{For $L \in \py, D \in \py$:}
+
+xxx need to recheck this
+
+$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$. OK.
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+
+Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
+
+Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
+
+But by Tip Merge, $\baseof{R} \ge \baseof{L}$
+
\end{document}