-Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
-This involves considering $D \in \py$.
+$$
+\begin{cases}
+ L \nothaspatch \p \land R \nothaspatch \p : & C \nothaspatch \p \\
+ L \haspatch \p \land R \haspatch \p : & C \haspatch \p \\
+ \text{otherwise} \land M \haspatch \p : & C \nothaspatch \p \\
+ \text{otherwise} \land M \nothaspatch \p : & C \haspatch \p
+\end{cases}
+$$
+\proofstarts
+~ Consider $D \in \py$.
\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 Own Contents for $L$).
So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
\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 Own Contents for $L$).
So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
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
if $M \haspatch \p$, $D \not\isin C$,
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
if $M \haspatch \p$, $D \not\isin C$,
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.