chiark / gitweb /
index 023a18a..01297a8 100644 (file)
@@ -65,8 +65,8 @@
\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}{%
@@ -81,7 +81,8 @@
\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}

\newcommand{\qed}{\square}
-\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+\newcommand{\proofstarts}{{\it Proof:}}
+\newcommand{\proof}[1]{\proofstarts #1 $\qed$}

\newcommand{\gathbegin}{\begin{gather} \tag*{}}
\newcommand{\gathnext}{\\ \tag*{}}
@@ -205,7 +206,7 @@ Ie, the two limbs of the RHS of Tip Contents are mutually exclusive.
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 \\ @@ -262,7 +263,7 @@ XXX proof TBD. \subsection{No Replay for Merge Results} -If we are constructing C, given +If we are constructing C, with, \gathbegin \mergeof{C}{L}{M}{R} \gathnext @@ -270,7 +271,7 @@ If we are constructing C, given \gathnext R \le C \end{gather} -No Replay is preserved. {\it Proof:} +No Replay is preserved. \proofstarts \subsubsection{For D=C:} D \isin C, D \le C. OK. @@ -424,7 +425,7 @@ Merge Results applies. \qed \subsection{Desired Contents} \[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C$
-{\it Proof.}
+\proofstarts

\subsubsection{For $D = C$:}

@@ -473,6 +474,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
\gathnext
\mergeof{C}{L}{M}{R}
\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.

\subsection{Conditions}

@@ -486,6 +488,21 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
\text{otherwise} : & \false
\end{cases}
}\]
+$\eqn{ Removal Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \haspatch \p + \implies + \pendsof{Y}{\py} = \pendsof{M}{\py} +}$
+$\eqn{ Addition Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \nothaspatch \p + \implies \left[ + \bigforall_{E \in \pendsof{X}{\py}} E \le Y + \right] +}$

\subsection{No Replay}

@@ -527,8 +544,76 @@ $\qed$

\subsection{Coherence and patch inclusion}

-xxx tbd
+Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
+This involves considering $D \in \py$.
+
+\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
+$D \not\isin L \land D \not\isin R$.  $C \not\in \py$ (otherwise $L +\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch).  So $D \neq C$.
+Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+
+\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
+$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
+(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
+
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
+
+For $D \neq C$: $D \le C \equiv D \le L \lor D \le R + \equiv D \isin L \lor D \isin R$.
+(Likewise $D \le C \equiv D \le X \lor D \le Y$.)
+
+Consider $D \neq C, D \isin X \land D \isin Y$:
+By $\merge$, $D \isin C$.  Also $D \le X$
+so $D \le C$.  OK for $C \haspatch \p$.
+
+Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
+By $\merge$, $D \not\isin C$.
+And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
+OK for $C \haspatch \p$.
+
+Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
+$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
+Thus by $\merge$, $D \isin C$.  And $D \le Y$ so $D \le C$.
+OK for $C \haspatch \p$.
+
+So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.

-xxx need to finish merge
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$C \haspatch \p \equiv M \nothaspatch \p$.
+
+\proofstarts
+
+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.
+
+$\qed$

\end{document}