traversal: Prove Recreate Base Beginning - Create Acyclic

[topbloke-formulae.git] / trav-proofs.tex

diff --git a/trav-proofs.tex b/trav-proofs.tex
index 3c2a1754cd8991bcc8cd1c4877812297cc9cdbd4..0a2d49127eb06a1d7c46e3ae8ad028955d27dd29 100644 (file)
--- a/trav-proofs.tex
+++ b/trav-proofs.tex
@@ -8,14 +8,16 @@ WIP WHAT ABOUT PROVING ALL THE TRAVERSAL RESULTS
 \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}
@@ -30,9 +32,16 @@ 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}
 
-$L = \tipdy$ 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. $L \nothaspatch \pq$. $\qed$
 
 \subsection{Tip Base Merge}