}\]
xxx proof tbd
+\[ \eqn{Ingredients Prevent Replay:}{
+ \left[
+ {C \hasparents \set A} \land
+ \\
+ \left(
+ D \isin C \implies
+ D = C \lor
+ \Largeexists_{A \in \set A} D \isin A
+ \right)
+ \right] \implies \left[
+ D \isin C \implies D \le C
+ \right]
+}\]
+\proof{
+ Trivial for $D = C$. Consider some $D \neq C$, $D \isin C$.
+ By the preconditions, there is some $A$ s.t. $D \in \set A$
+ and $D \isin A$. By No Replay for $A$, $D \le A$. And
+ $A \le C$ so $D \le C$.
+}
+
\[ \eqn{Totally Foreign Contents:}{
\bigforall_{C \hasparents \set A}
\left[
\right]
\implies
\left[
- D \isin C
+ D \le C
\implies
\patchof{D} = \bot
\right]
}\]
-xxx proof tbd
-
-\subsection{No Replay for Merge Results}
-
-If we are constructing $C$, with,
-\gathbegin
- \mergeof{C}{L}{M}{R}
-\gathnext
- L \le C
-\gathnext
- R \le C
-\end{gather}
-No Replay is preserved. \proofstarts
-
-\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.
-
-\subsubsection{For $D \isin L \land D \isin R$:}
-$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK.
-
-\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
-$D \not\isin C$. OK.
-
-\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \not\isin M$:}
-$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
-R$ so $D \le C$. OK.
-
-\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \isin M$:}
-$D \not\isin C$. OK.
-
-$\qed$
+\proof{
+Consider some $D \le C$. If $D = C$, $\patchof{D} = \bot$ trivially.
+If $D \neq C$ then $D \le A$ where $A \in \set A$. By Foreign
+Contents of $A$, $\patchof{D} = \bot$.
+}
\section{Commit annotation}
Topbloke strips the metadata when exporting.
\subsection{No Replay}
-Trivial.
+
+Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
If $A, C \in \py$ then by Calculation of Ends for
\subsection{Foreign Contents:}
-Only relevant if $\patchof{C} = \bot$. Trivial by Foreign Contents of
-$A$. $\qed$
-
-xxx fixme not trivial use Totally Foreign Contents
+Only relevant if $\patchof{C} = \bot$, and in that case Totally
+Foreign Contents applies. $\qed$
\section{Create Base}
\gathbegin
B \hasparents \{ L \}
\gathnext
- \patchof{B} = \pa{B}
+ \patchof{B} = \pan{Q}
\gathnext
D \isin B \equiv D \isin L \lor D = B
\end{gather}
\patchof{L} = \pa{L} \lor \patchof{L} = \bot
}\]
\[ \eqn{ Non-recursion }{
- L \not\in \pa{B}
+ L \not\in \pa{Q}
}\]
\subsection{No Replay}
-If $\patchof{L} = \pa{L}$, trivial by Base Acyclic for $L$.
-
-If $\patchof{L} = \bot$, xxx
-
-Trivial from Base Acyclic for $L$. $\qed$
+Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
-Not applicable. $\qed$
+Not applicable.
\subsection{Tip Contents}
-Not applicable. $\qed$
+Not applicable.
\subsection{Base Acyclic}
-xxx
+Consider some $D \isin B$. If $D = B$, $D \in \pn$, OK.
+
+If $D \neq B$, $D \isin L$. By No Replay of $D$ in $L$, $D \le L$.
+Thus by Foreign Contents of $L$, $\patchof{D} = \bot$. OK.
-xxx unfinished
+xxx this is wrong
+
+$\qed$
+
+\subsection{Coherence and Patch Inclusion}
+
+Consider some $D \in \p$.
+$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$.
+
+Thus $L \haspatch \p \implies B \haspatch P$
+and $L \nothaspatch \p \implies B \nothaspatch P$.
+
+$\qed$.
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq B$
+so $D \isin B \equiv D \isin L$.
+By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le B$.
+So $D \isin B \equiv D \le B$. $\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
\section{Create Tip}
-xxx tbd
+Given a Topbloke base $B$, create a tip branch initial commit B.
+\gathbegin
+ C \hasparents \{ B \}
+\gathnext
+ \patchof{B} = \pay{Q}
+\gathnext
+ D \isin C \equiv D \isin B \lor D = C
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{B} = \pan{Q}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Trivially, $\pendsof{C}{\pan{Q}} = \{B\}$ so $\baseof{C} = B$.
+
+\subsection{Tip Contents}
+
+Consider some arbitrary commit $D$. If $D = C$, trivially satisfied.
+
+If $D \neq C$, $D \isin C \equiv D \isin B$.
+By Base Acyclic of $B$, $D \isin B \implies D \not\in \pay{Q}$.
+So $D \isin C \equiv D \isin \baseof{B}$.
+
+$\qed$
+
+xxx up to here
\section{Anticommit}
L \haspatch \pry
}\]
-\subsection{Ordering of ${L, R^+, R^-}$:}
+\subsection{Ordering of Ingredients:}
By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$.
\subsection{No Replay}
-No Replay for Merge Results applies. $\qed$
+By definition of $\merge$,
+$D \isin C \implies D \isin L \lor D \isin R^- \lor D = C$.
+So, by Ordering of Ingredients,
+Ingredients Prevent Replay applies. $\qed$
\subsection{Desired Contents}
$\qed$
-\subsection{Foreign Contents:}
+\subsection{Foreign Contents}
-Not applicable. $\qed$
+Not applicable.
\section{Merge}
\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
+\[ \eqn{ Foreign Merges }{
+ \patchof{L} = \bot \equiv \patchof{R} = \bot
+}\]
\subsection{Non-Topbloke merges}
-We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$.
+We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$
+(Foreign Merges, above).
I.e. not only is it forbidden to merge into a Topbloke-controlled
branch without Topbloke's assistance, it is also forbidden to
merge any Topbloke-controlled branch into any plain git branch.
Given those conditions, Tip Merge and Merge Acyclic do not apply.
And $Y \not\in \py$ so $\neg [ Y \haspatch \p ]$ so neither
-Merge Ends condition applies. Good.
+Merge Ends condition applies.
+
+So a plain git merge of non-Topbloke branches meets the conditions and
+is therefore consistent with our scheme.
\subsection{No Replay}
-No Replay for Merge Results applies. $\qed$
+By definition of $\merge$,
+$D \isin C \implies D \isin L \lor D \isin R \lor D = C$.
+So, by Ingredients,
+Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
$\qed$
-\subsection{Foreign Contents:}
-
-xxx use Totally Foreign Contents
+\subsection{Foreign Contents}
-If $\patchof{C} = \bot$, by Foreign Merges
-$\patchof{L} = \patchof{R} = \bot$.
+Only relevant if $\patchof{L} = \bot$, in which case
+$\patchof{C} = \bot$ and by Foreign Merges $\patchof{R} = \bot$,
+so Totally Foreign Contents applies. $\qed$
\end{document}