\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}{%
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{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
\bigforall_{C \in \py} D \isin C \equiv
\begin{cases}
D \in \py : & D \le C \\
\subsection{No Replay for Merge Results}
-If we are constructing $C$, given
+If we are constructing $C$, with,
\gathbegin
\mergeof{C}{L}{M}{R}
\gathnext
\gathnext
\mergeof{C}{L}{M}{R}
\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\subsection{Conditions}
\text{otherwise} : & \false
\end{cases}
}\]
+\[ \eqn{ Removal Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \haspatch \p
+ \implies
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
+}\]
\subsection{No Replay}
\subsection{Coherence and patch inclusion}
-Need to determine $C \haspatch P$ based on $L,M,R \haspatch P$.
+Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
This involves considering $D \in \py$.
-We will use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
-
-\subsubsection{For $L \nothaspatch P, R \nothaspatch P$:}
+\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
-\in \py$ ie $L \haspatch P$ by Tip Self Inpatch). So $D \neq C$.
-Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch P$.
+\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch). So $D \neq C$.
+Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
-\subsubsection{For $L \haspatch P, R \haspatch P$:}
+\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
-Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch P$.
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
\equiv D \isin L \lor D \isin R$.
Consider $D \neq C, D \isin X \land D \isin Y$:
By $\merge$, $D \isin C$. Also $D \le X$
-so $D \le C$. OK for $C \haspatch P$.
+so $D \le C$. OK for $C \haspatch \p$.
Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
By $\merge$, $D \not\isin C$.
And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
-OK for $C \haspatch P$.
+OK for $C \haspatch \p$.
Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$.
-OK for $C \haspatch P$.
+OK for $C \haspatch \p$.
+
+So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$C \haspatch \p \equiv M \nothaspatch \p$.
+
+\proofstarts
-So indeed $L \haspatch P \land R \haspatch P \implies C \haspatch P$.
+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
+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.
+
+$\qed$
\end{document}