\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 Ends }{
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
+\[ \eqn{ Removal Merge Ends }{
X \not\haspatch \p \land
Y \haspatch \p \land
- E \in \pendsof{X}{\py}
+ M \haspatch \p
\implies
- E \le Y
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
}\]
\subsection{No Replay}
\proofstarts
-Merge Ends applies. Recall that we are considering $D \in \py$.
+One of the Merge Ends conditions applies.
+Recall that we are considering $D \in \py$.
$D \isin Y \equiv D \le Y$. $D \not\isin X$.
We will show for each of
various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
(which suffices by definition of $\haspatch$ and $\nothaspatch$).
-Consider $D = C$. Thus $C \in \py, L \in \py$, and by Tip
+Consider $D = C$: Thus $C \in \py, L \in \py$, and by Tip
Self Inpatch $L \haspatch \p$, so $L=Y, R=X$. By Tip Merge,
$M=\baseof{L}$. So by Base Acyclic $D \not\isin M$, i.e.
$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK.
+Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+$D \le Y$ so $D \le C$.
+$D \not\isin M$ so by $\merge$, $D \isin C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
+$D \not\le Y$. If $D \le X$ then
+$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
+Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
+Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \isin Y$:
+$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
+and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
+Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \not\isin Y$:
+By $\merge$, $D \not\isin C$. OK.
+
+$\qed$
+
+\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 need worry only about $C \in \py$. And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$. We will use the coherence and
+patch inclusion of $C$ as just proved.
+
+Firstly we prove $C \haspatch \p$: If $R \in \py$, this is true by
+coherence/inclusion $C \haspatch \p$. So by definition of
+$\haspatch$, $D \isin C \equiv D \le C$. OK.
+
+\subsubsection{For $L \in \py, D \in \py, $:}
+$R \haspatch \p$ so
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+
+
+%D \in \py$:}
+
+
+
+xxx the coherence is not that useful ?
+
+$L \haspatch \p$ by
+
+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}