\text{otherwise} : & \false
\end{cases}
}\]
-\[ \eqn{ Merge Ends }{
+\[ \eqn{ Removal Merge Ends }{
X \not\haspatch \p \land
- Y \haspatch \p
- \implies \left[
- \bigforall_{E \in \pendsof{X}{\py}}
- E \le Y
- \right]
+ Y \haspatch \p \land
+ M \haspatch \p
+ \implies
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
}\]
\subsection{No Replay}
\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 $D \isin C \equiv M \nothaspatch \p \land D \le C$
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 Merge Ends and
+$D \in \pancsof{X}{\py}$, 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}