\subsection{Traversal Lemmas}
Firstly, some lemmas.
-
\statement{Tip Correct Contents}{
- \tipcy \haspatch \pd
+ \tipcy \haspatch \pa E
\equiv
- \pc = \pd \lor \pc \hasdep \pd
+ \pa E = \pc \lor \pa E \isdep \pc
}
\proof{
- WIP
+ For $\pc = \pa E$, Tip Own Contents suffices.
+ For $\pc \neq \pa E$, Exclusive Tip Contents
+ gives $D \isin \tipcy \equiv D \isin \baseof{\tipcy}$
+ which by Correct Base $\equiv D \isin \tipcn$.
}
\subsection{Base Dependency Merge, Base Sibling Merge}
\subsection{Recreate Base Beginning}
+To recap we are executing Create Base with
+$L = \tipdy$ and $\pq = \pc$.
+
\subsubsection{Create Acyclic}
-$L = \tipdy$ so
+By Tip Correct Contents of $L$,
+$L \haspatch \pa E \equiv \pa E = \pd \lor \pa E \isdep \pd$.
+Now $\pd \isdirdep \pc$,
+so by Coherence, and setting $\pa E = \pc$,
+$L \nothaspatch \pc$. I.e. $L \nothaspatch \pq$. OK.
+
+That's everything for Create Base. $\qed$
+
+\subsection{Recreate Base Final Declaration}
+
+\subsubsection{Base Only} $\patchof{W} = \patchof{L} = \pn$. OK.
+
+\subsubsection{Unique Tips}
+
+Want to prove that for any $\p \isin C$, $\tipdy$ is a suitable $T$.
+
+WIP
\subsection{Tip Base Merge}