+WIP WHAT ABOUT PROVING ALL THE TRAVERSAL RESULTS
+
+\subsection{Traversal Lemmas}
+
+Firstly, some lemmas.
+\statement{Tip Correct Contents}{
+ \tipcy \haspatch \pa E
+ \equiv
+ \pa E = \pc \lor \pa E \isdep \pc
+}
+\proof{
+ 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}
+
+We do not prove that the preconditions are met. Instead, we check
+them at runtime. If they turn out not to be met, we abandon
+\alg{Merge-Base} and resort to \alg{Recreate-Base}.
+
+TODO COMPLETE MERGE-BASE STUFF
+
+WIP WHAT ABOUT PROVING ALL THE TRAVERSAL RESULTS
+
+\subsection{Recreate Base Beginning}
+
+To recap we are executing Create Base with
+$L = \tipdy$ and $\pq = \pc$.
+
+\subsubsection{Create Acyclic}
+
+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
+