\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
\newcommand{\patchof}[1]{\patch ( #1 ) }
\newcommand{\baseof}[1]{\base ( #1 ) }
+\newcommand{\eqntag}[2]{ #2 \tag*{\mbox{#1}} }
\newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
-\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
%\newcommand{\bigforall}{\mathop{\hbox{\huge$\forall$}}}
\newcommand{\bigforall}{%
are respectively the base and tip git branches. $\p$ may be used
where the context requires a set, in which case the statement
is to be taken as applying to both $\py$ and $\pn$.
-All these sets are distinct. Hence:
+None of these sets overlap. Hence:
\item[ $ \patchof{ C } $ ]
Either $\p$ s.t. $ C \in \p $, or $\bot$.
\end{cases}
$
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \equiv D \isin R : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin M \equiv D \isin L : & D = C \lor D \isin R \\
+ D \isin M \equiv D \isin R : & D = C \lor D \isin L \\
+ \end{cases}
+$
+
\end{basedescript}
\newpage
\section{Invariants}
Let $B = \baseof{C}$ in $D \isin \baseof{C}$. Now $B \in \pn$.
So by Base Acyclic $D \isin B \implies D \notin \py$.
}
-\[ \corrolary{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
\bigforall_{C \in \py} D \isin C \equiv
\begin{cases}
D \in \py : & D \le C \\
\begin{cases}
R \in \py : & \baseof{R} \ge \baseof{L}
\land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
- R \in \pn : & R \ge \baseof{L}
- \land M = \baseof{L} \\
+ R \in \pn : & M = \baseof{L} \\
\text{otherwise} : & \false
\end{cases}
}\]
-\[ \eqn{ Merge Ends }{
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
+\[ \eqn{ Removal Merge Ends }{
X \not\haspatch \p \land
Y \haspatch \p \land
- E \in \pendsof{X}{\py}
+ M \haspatch \p
\implies
- E \le Y
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
}\]
\subsection{No Replay}
\subsubsection{For $R \in \pn$:}
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
$A \le \baseof{L} \implies A \le R$, so
$A \le R \lor A \le \baseof{L} \equiv A \le R$.
Thus $A \le C \equiv A \le R$.
$\qed$
-\subsection{Coherence and patch inclusion}
+\subsection{Coherence and Patch Inclusion}
Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
This involves considering $D \in \py$.
\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
-$C \haspatch \p \equiv C \nothaspatch M$.
+$C \haspatch \p \equiv M \nothaspatch \p$.
\proofstarts
-Merge Ends applies.
+One of the Merge Ends conditions applies.
+Recall that we are considering $D \in \py$.
+$D \isin Y \equiv D \le Y$. $D \not\isin X$.
+We will show for each of
+various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
+(which suffices by definition of $\haspatch$ and $\nothaspatch$).
+
+Consider $D = C$: Thus $C \in \py, L \in \py$, and by Tip
+Self Inpatch $L \haspatch \p$, so $L=Y, R=X$. By Tip Merge,
+$M=\baseof{L}$. So by Base Acyclic $D \not\isin M$, i.e.
+$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+$D \le Y$ so $D \le C$.
+$D \not\isin M$ so by $\merge$, $D \isin C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
+$D \not\le Y$. If $D \le X$ then
+$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
+Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
+Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \isin Y$:
+$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
+and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
+Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \not\isin Y$:
+By $\merge$, $D \not\isin C$. OK.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
+R$. And $D \neq C$. So $D \not\isin C$. $\qed$
+
+\subsection{Tip Contents}
+
+We need worry only about $C \in \py$.
+And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
+of $C$, and its Coherence and Patch Inclusion, as just proved.
+
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by Coherence/Inclusion $C \haspatch \p$ . If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$. So by Coherence/Inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
+
+We will consider an arbitrary commit $D$
+and prove the Exclusive Tip Contents form.
+
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$. OK.
+
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
+
+$D \neq C$. By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+xxx up to here
+
+%D \in \py$:}
+
+
+
+xxx the coherence is not that useful ?
+
+$L \haspatch \p$ by
+
+xxx need to recheck this
+
+$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$. OK.
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
-$D \isin Y \equiv D \le Y$. $D \not\isin X$. Recall that we
-are considering $D \in \py$.
+Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
-Consider $D = C$. Thus $C \in \py, L \in \py$.
-But $X \not\haspatch \p$ means xxx wip
-But $X \not\haspatch \p$ means $D \not\in X$,
+Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
-so we have $L = Y, R =
-X$. Thus $R \not\haspatch \p$ and by Tip Self Inpatch $R \not\in
-\py$. Thus by Tip Merge $R \in \pn$ and $M = \baseof{L}$.
-So by Base Acyclic, $M \nothaspatch \py$. Thus we are expecting
-$C \haspatch \py$. And indeed $D \isin C$ and $D \le C$. OK.
+But by Tip Merge, $\baseof{R} \ge \baseof{L}$
\end{document}
~ian / topbloke-formulae.git / blobdiff