\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
+\[ \eqn{ Suitable Tip }{
+ \bigexists_T
+ \pendsof{J}{\py} = \{ T \}
+ \land
+ \forall_{E \in \pendsof{K}{\py}} T \ge E
+ , \text{where} \{J,K\} = \{L,R\}
+}\]
\[ \eqn{ Foreign Merges }{
\patchof{L} = \bot \implies \patchof{R} = \bot
}\]
\subsection{Coherence and Patch Inclusion}
-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. $C \nothaspatch \p$.
+OK.
\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 (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
-$M \haspatch \p \implies C \nothaspatch \p$.
-$M \nothaspatch \p \implies C \haspatch \p$.
-
-\proofstarts
-
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 $M \haspatch \p$, $D \not\isin C$,
-whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
+whereas if $M \nothaspatch \p$, $D \isin C \equiv D \le C$.
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.
Consider $D = C$: Thus $C \in \py, L \in \py$.
-By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
+By Tip Own Contents, $L \haspatch \p$ so $L \neq X$,
therefore we must have $L=Y$, $R=X$.
-By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
+Conversely $R \not\in \py$
+so by Tip Merge $M = \baseof{L}$. Thus $M \in \pn$ so
by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
and $D \le C$. OK.
$\qed$
+\subsection{Unique Tips}
+
+For $L \in \py$, trivially $\pendsof{C}{\py} = C$ so $T = C$ is
+suitable.
+
+For $L \not\in \py$, $\pancsof{C}{\py} = \pancsof{L}{\py} \cup
+\pancsof{R}{\py}$. So $T$ from Suitable Tip is a suitable $T$ for
+Unique Tips.
+
+$\qed$
+
\subsection{Foreign Inclusion}
Consider some $D$ s.t. $\patchof{D} = \bot$.