\usepackage{mdwlist}
%\usepackage{accents}
+\usepackage{fancyhdr}
+\pagestyle{fancy}
+\lhead[\rightmark]{}
+
+\let\stdsection\section
+\renewcommand\section{\newpage\stdsection}
+
\renewcommand{\ge}{\geqslant}
\renewcommand{\le}{\leqslant}
\newcommand{\nge}{\ngeqslant}
\section{Some lemmas}
-\[ \eqn{Alternative (overlapping) formulations defining
- $\mergeof{C}{L}{M}{R}$:}{
+\subsection{Alternative (overlapping) formulations of $\mergeof{C}{L}{M}{R}$}
+$$
D \isin C \equiv
\begin{cases}
D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
D \isin L \nequiv D \isin M : & D = C \lor D \isin L \\
\text{as above with L and R exchanged}
\end{cases}
-}\]
-\proof{
- Truth table xxx
-
- Original definition is symmetrical in $L$ and $R$.
+$$
+\proof{ ~ Truth table (ordered by original definition): \\
+ \begin{tabular}{cccc|c|cc}
+ $D = C$ &
+ $\isin L$ &
+ $\isin M$ &
+ $\isin R$ & $\isin C$ &
+ $L$ vs. $R$ & $L$ vs. $M$
+ \\\hline
+ y & ? & ? & ? & y & ? & ? \\
+ n & y & y & y & y & $\equiv$ & $\equiv$ \\
+ n & y & n & y & y & $\equiv$ & $\nequiv$ \\
+ n & n & y & n & n & $\equiv$ & $\nequiv$ \\
+ n & n & n & n & n & $\equiv$ & $\equiv$ \\
+ n & y & y & n & n & $\nequiv$ & $\equiv$ \\
+ n & n & y & y & n & $\nequiv$ & $\nequiv$ \\
+ n & y & n & n & y & $\nequiv$ & $\nequiv$ \\
+ n & n & n & y & y & $\nequiv$ & $\equiv$ \\
+ \end{tabular} \\
+ And original definition is symmetrical in $L$ and $R$.
}
-\[ \eqn{Exclusive Tip Contents:}{
+\subsection{Exclusive Tip Contents}
+$$
\bigforall_{C \in \py}
\neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C )
\Bigr]
-}\]
+$$
Ie, the two limbs of the RHS of Tip Contents are mutually exclusive.
\proof{
\end{cases}
}\]
-\[ \eqn{Tip Self Inpatch:}{
+\subsection{Tip Self Inpatch}
+$$
\bigforall_{C \in \py} C \haspatch \p
-}\]
+$$
Ie, tip commits contain their own patch.
\proof{
D \isin C \equiv D \le C $
}
-\[ \eqn{Exact Ancestors:}{
+\subsection{Exact Ancestors}
+$$
\bigforall_{ C \hasparents \set{R} }
D \le C \equiv
( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
\lor D = C
-}\]
-xxx proof tbd
+$$
+\proof{ ~ Trivial.}
-\[ \eqn{Transitive Ancestors:}{
+\subsection{Transitive Ancestors}
+$$
\left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv
\left[ \bigforall_{ A \in \pancsof{C}{\set P} } A \le M \right]
-}\]
+$$
\proof{
The implication from right to left is trivial because
by the LHS. And $A \le A''$.
}
-\[ \eqn{Calculation Of Ends:}{
+\subsection{Calculation Of Ends}
+$$
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
\begin{cases}
\left\{ E \Big|
\Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
- \Bigl[ \Largenexists_{B \in \set A}
- E \neq B \land E \le B \Bigr]
+ \Bigl[ \Largenexists_{B \in \set A, F \in \pendsof{B}{\p}}
+ E \neq F \land E \le F \Bigr]
\right\}
\end{cases}
-}\]
-xxx proof tbd
+$$
+\proof{
+Trivial for $C \in \set P$. For $C \not\in \set P$,
+$\pancsof{C}{\set P} = \bigcup_{A \in \set A} \pancsof{A}{\set P}$.
+So $\pendsof{C}{\set P} \subset \bigcup_{E in \set E} \pendsof{E}{\set P}$.
+Consider some $E \in \pendsof{A}{\set P}$. If $\exists_{B,F}$ as
+specified, then either $F$ is going to be in our result and
+disqualifies $E$, or there is some other $F'$ (or, eventually,
+an $F''$) which disqualifies $F$.
+Otherwise, $E$ meets all the conditions for $\pends$.
+}
-\[ \eqn{Ingredients Prevent Replay:}{
+\subsection{Ingredients Prevent Replay}
+$$
\left[
{C \hasparents \set A} \land
\\
\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. $D \in \set A$
$A \le C$ so $D \le C$.
}
-\[ \eqn{Simple Foreign Inclusion:}{
+\subsection{Simple Foreign Inclusion}
+$$
\left[
C \hasparents \{ L \}
\land
\bigforall_{D} D \isin C \equiv D \isin L \lor D = C
\right]
\implies
+ \left[
\bigforall_{D \text{ s.t. } \patchof{D} = \bot}
D \isin C \equiv D \le C
-}\]
+ \right]
+$$
\proof{
Consider some $D$ s.t. $\patchof{D} = \bot$.
If $D = C$, trivially true. For $D \neq C$,
So $D \isin C \equiv D \le C$.
}
-\[ \eqn{Totally Foreign Contents:}{
+\subsection{Totally Foreign Contents}
+$$
\bigforall_{C \hasparents \set A}
\left[
\patchof{C} = \bot \land
\implies
\patchof{D} = \bot
\right]
-}\]
+$$
\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
A simple single-parent forward commit $C$ as made by git-commit.
\begin{gather}
-\tag*{} C \hasparents \{ A \} \\
-\tag*{} \patchof{C} = \patchof{A} \\
-\tag*{} D \isin C \equiv D \isin A \lor D = C
+\tag*{} C \hasparents \{ L \} \\
+\tag*{} \patchof{C} = \patchof{L} \\
+\tag*{} D \isin C \equiv D \isin L \lor D = C
\end{gather}
This also covers Topbloke-generated commits on plain git branches:
Topbloke strips the metadata when exporting.
Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
-If $A, C \in \py$ then by Calculation of Ends for
+If $L, C \in \py$ then by Calculation of Ends for
$C, \py, C \not\in \py$:
-$\pendsof{C}{\pn} = \pendsof{A}{\pn}$ so
-$\baseof{C} = \baseof{A}$. $\qed$
+$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ so
+$\baseof{C} = \baseof{L}$. $\qed$
\subsection{Tip Contents}
-We need to consider only $A, C \in \py$. From Tip Contents for $A$:
-\[ D \isin A \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \]
+We need to consider only $L, C \in \py$. From Tip Contents for $L$:
+\[ D \isin L \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L ) \]
Substitute into the contents of $C$:
-\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A )
+\[ D \isin C \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L )
\lor D = C \]
Since $D = C \implies D \in \py$,
and substituting in $\baseof{C}$, this gives:
\[ D \isin C \equiv D \isin \baseof{C} \lor
- (D \in \py \land D \le A) \lor
+ (D \in \py \land D \le L) \lor
(D = C \land D \in \py) \]
\[ \equiv D \isin \baseof{C} \lor
- [ D \in \py \land ( D \le A \lor D = C ) ] \]
+ [ D \in \py \land ( D \le L \lor D = C ) ] \]
So by Exact Ancestors:
\[ D \isin C \equiv D \isin \baseof{C} \lor ( D \in \py \land D \le C
) \]
\subsection{Base Acyclic}
-Need to consider only $A, C \in \pn$.
+Need to consider only $L, C \in \pn$.
For $D = C$: $D \in \pn$ so $D \not\in \py$. OK.
-For $D \neq C$: $D \isin C \equiv D \isin A$, so by Base Acyclic for
-$A$, $D \isin C \implies D \not\in \py$.
+For $D \neq C$: $D \isin C \equiv D \isin L$, so by Base Acyclic for
+$L$, $D \isin C \implies D \not\in \py$.
$\qed$
Need to consider $D \in \py$
-\subsubsection{For $A \haspatch P, D = C$:}
+\subsubsection{For $L \haspatch P, D = C$:}
Ancestors of $C$:
$ D \le C $.
Contents of $C$:
$ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $.
-\subsubsection{For $A \haspatch P, D \neq C$:}
-Ancestors: $ D \le C \equiv D \le A $.
+\subsubsection{For $L \haspatch P, D \neq C$:}
+Ancestors: $ D \le C \equiv D \le L $.
-Contents: $ D \isin C \equiv D \isin A \lor f $
-so $ D \isin C \equiv D \isin A $.
+Contents: $ D \isin C \equiv D \isin L \lor f $
+so $ D \isin C \equiv D \isin L $.
So:
-\[ A \haspatch P \implies C \haspatch P \]
+\[ L \haspatch P \implies C \haspatch P \]
-\subsubsection{For $A \nothaspatch P$:}
+\subsubsection{For $L \nothaspatch P$:}
-Firstly, $C \not\in \py$ since if it were, $A \in \py$.
+Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
-Now by contents of $A$, $D \notin A$, so $D \notin C$.
+Now by contents of $L$, $D \notin L$, so $D \notin C$.
So:
-\[ A \nothaspatch P \implies C \nothaspatch P \]
+\[ L \nothaspatch P \implies C \nothaspatch P \]
$\qed$
\subsection{Foreign Inclusion:}
~ian / topbloke-formulae.git / blobdiff