X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=trav-proofs.tex;h=e69de29bb2d1d6434b8b29ae775ad8c2e48c5391;hp=0a2d49127eb06a1d7c46e3ae8ad028955d27dd29;hb=6e7093d49c00150e48ebc264fb1a29de2004aeed;hpb=63b8fab3f0b73ade21eaa412eeaed13aeb5ee691

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-\section{Traversal phase --- proofs}
-
-For each operation called for by the traversal algorithms, we prove
-that the commit generation preconditions are met.
-
-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. $L \nothaspatch \pq$. $\qed$
-
-\subsection{Tip Base Merge}
-
-$L = W$, $R = \tipcn$.
-
-TODO TBD
-
-Afterwards, $\baseof{W} = \tipcn$.
-
-\subsection{Tip Source Merge}
-
-In fact, we do this backwards: $L = S$, $R = W$.  Since $S \in \pcy$,
-the resulting $C \in \pcy$ and the remaining properties of the Merge
-commit construction are symmetrical in $L$ and $R$ so this is fine.
-
-By the results of Tip Base Merge, $\baseof{W} = \tipcn$.
-
-By Base Ends Supreme, $\tipcn \ge \baseof{S}$ i.e.
-$\baseof{R} \ge \baseof{L}$.
-
-Either $\baseof{L} = \baseof{M}$, or we must choose a different $M$ in
-which case $M = \baseof{S}$ will suffice.
-