}\]
\[ \eqn{ Merge Ends }{
X \not\haspatch \p \land
- Y \haspatch \p \land
- E \in \pendsof{X}{\py}
- \implies
+ Y \haspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}}
E \le Y
+ \right]
}\]
\subsection{No Replay}
\proofstarts
-Merge Ends applies.
+Merge Ends 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 $\nothaspatch$).
-$D \isin Y \equiv D \le Y$. $D \not\isin X$. Recall that we
-are considering $D \in \py$.
-
-Consider $D = C$. Thus $C \in \py, L \in \py$.
-But $X \not\haspatch \p$ means xxx wip
-But $X \not\haspatch \p$ means $D \not\in X$,
-
-so we have $L = Y, R =
-X$. Thus $R \not\haspatch \p$ and by Tip Self Inpatch $R \not\in
-\py$. Thus by Tip Merge $R \in \pn$ and $M = \baseof{L}$.
-So by Base Acyclic, $M \nothaspatch \py$. Thus we are expecting
-$C \haspatch \py$. And indeed $D \isin C$ and $D \le C$. OK.
+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.
\end{document}