\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}
\newcommand{\py}{\pay{P}}
\newcommand{\pn}{\pan{P}}
+\newcommand{\pq}{\pa{Q}}
+\newcommand{\pqy}{\pay{Q}}
+\newcommand{\pqn}{\pan{Q}}
+
\newcommand{\pr}{\pa{R}}
\newcommand{\pry}{\pay{R}}
\newcommand{\prn}{\pan{R}}
\item[ $ C \ge D $ ]
$C$ is a descendant of $D$ in the git commit
-graph. This is a partial order, namely the transitive closure of
+graph. This is a partial order, namely the transitive closure of
$ D \in \set X $ where $ C \hasparents \set X $.
\item[ $ C \has D $ ]
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 $\pn$.
-None of these sets overlap. Hence:
+All of these sets are disjoint. Hence:
\item[ $ \patchof{ C } $ ]
-Either $\p$ s.t. $ C \in \p $, or $\bot$.
+Either $\p$ s.t. $ C \in \p $, or $\bot$.
A function from commits to patches' sets $\p$.
\item[ $ \pancsof{C}{\set P} $ ]
-$ \{ A \; | \; A \le C \land A \in \set P \} $
+$ \{ A \; | \; A \le C \land A \in \set P \} $
i.e. all the ancestors of $C$
which are in $\set P$.
\item[ $ \pendsof{C}{\set P} $ ]
$ \{ E \; | \; E \in \pancsof{C}{\set P}
\land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}}
- E \neq A \land E \le A \} $
+ E \neq A \land E \le A \} $
i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$.
\item[ $ \baseof{C} $ ]
\item[ $ C \nothaspatch \p $ ]
$\displaystyle \bigforall_{D \in \py} D \not\isin C $.
-~ Informally, $C$ has none of the contents of $\p$.
+~ Informally, $C$ has none of the contents of $\p$.
-Non-Topbloke commits are $\nothaspatch \p$ for all $\p$. This
+Commits on Non-Topbloke branches are $\nothaspatch \p$ for all $\p$. This
includes commits on plain git branches made by applying a Topbloke
patch. If a Topbloke
patch is applied to a non-Topbloke branch and then bubbles back to
-the relevant Topbloke branches, we hope that
+the relevant Topbloke branches, we hope that
if the user still cares about the Topbloke patch,
git's merge algorithm will DTRT when trying to re-apply the changes.
(D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
\text{otherwise} : & D \not\isin M
\end{cases}
-$
+$
\end{basedescript}
\newpage
\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:}{
- \bigforall_{C \in \py}
+\subsection{Exclusive Tip Contents}
+Given Base Acyclic for $C$,
+$$
+ \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}
+Given Exclusive Tip Contents and Base Acyclic for $C$,
+$$
\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} }
+ \left[
D \le C \equiv
( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
\lor D = C
-}\]
-xxx proof tbd
+ \right]
+$$
+\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
$ \pends() \subset \pancs() $.
-For the implication from left to right:
+For the implication from left to right:
by the definition of $\mathcal E$,
for every such $A$, either $A \in \pends()$ which implies
$A \le M$ by the LHS directly,
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}
\\
C \not\in \p : & \displaystyle
\left\{ E \Big|
- \Bigl[ \Largeexists_{A \in \set A}
+ \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{Totally Foreign Contents:}{
- \bigforall_{C \hasparents \set A}
+\subsection{Ingredients Prevent Replay}
+$$
+ \left[
+ {C \hasparents \set A} \land
+ \\
+ \bigforall_{D}
+ \left(
+ D \isin C \implies
+ D = C \lor
+ \Largeexists_{A \in \set A} D \isin A
+ \right)
+ \right] \implies \left[ \bigforall_{D}
+ D \isin C \implies D \le C
+ \right]
+$$
+\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 $D \le C$.
+}
+
+\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 $D \neq C$,
+by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
+
+\subsection{Totally Foreign Contents}
+$$
\left[
+ C \hasparents \set A \land
\patchof{C} = \bot \land
\bigforall_{A \in \set A} \patchof{A} = \bot
\right]
\implies
\left[
+ \bigforall_{D}
D \le C
\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
Contents of $A$, $\patchof{D} = \bot$.
}
-\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.
-
-$\qed$
-
\section{Commit annotation}
We annotate each Topbloke commit $C$ with:
\gathnext
\baseof{C}, \text{ if } C \in \py
\gathnext
- \bigforall_{\pa{Q}}
- \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
+ \bigforall_{\pq}
+ \text{ either } C \haspatch \pq \text{ or } C \nothaspatch \pq
\gathnext
- \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+ \bigforall_{\pqy \not\ni C} \pendsof{C}{\pqy}
\end{gather}
$\patchof{C}$, for each kind of Topbloke-generated commit, is stated
Whether $\baseof{C}$ is required, and if so what the value is, is
stated in the proof of Unique Base for each kind of commit.
-$C \haspatch \pa{Q}$ or $\nothaspatch \pa{Q}$ is represented as the
-set $\{ \pa{Q} | C \haspatch \pa{Q} \}$. Whether $C \haspatch \pa{Q}$
+$C \haspatch \pq$ or $\nothaspatch \pq$ is represented as the
+set $\{ \pq | C \haspatch \pq \}$. Whether $C \haspatch \pq$
is in stated
-(in terms of $I \haspatch \pa{Q}$ or $I \nothaspatch \pa{Q}$
-for the ingredients $I$),
+(in terms of $I \haspatch \pq$ or $I \nothaspatch \pq$
+for the ingredients $I$)
in the proof of Coherence for each kind of commit.
-$\pendsof{C}{\pa{Q}^+}$ is computed, for all Topbloke-generated commits,
+$\pendsof{C}{\pq^+}$ is computed, for all Topbloke-generated commits,
using the lemma Calculation of Ends, above.
We do not annotate $\pendsof{C}{\py}$ for $C \in \py$. Doing so would
make it wrong to make plain commits with git because the recorded $\pends$
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.
\subsection{No Replay}
-Trivial.
+
+Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
-If $A, C \in \py$ then by Calculation of Ends for
-$C, \py, C \not\in \py$:
-$\pendsof{C}{\pn} = \pendsof{A}{\pn}$ so
-$\baseof{C} = \baseof{A}$. $\qed$
+If $L, C \in \py$ then by Calculation of Ends,
+$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ 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:
+Since $D = C \implies D \in \py$,
+and substituting in $\baseof{C}$, from Unique Base above, 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:}
+\subsection{Foreign Inclusion:}
-If $D = C$, trivial. For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
+Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Contents:}
\section{Create Base}
-Given $L$, create a Topbloke base branch initial commit $B$.
+Given a starting point $L$ and a proposed patch $\pq$,
+create a Topbloke base branch initial commit $B$.
\gathbegin
B \hasparents \{ L \}
\gathnext
- \patchof{B} = \pan{B}
+ \patchof{B} = \pqn
\gathnext
D \isin B \equiv D \isin L \lor D = B
\end{gather}
\subsection{Conditions}
-\[ \eqn{ Ingredients }{
- \patchof{L} = \pa{L} \lor \patchof{L} = \bot
-}\]
-\[ \eqn{ Non-recursion }{
- L \not\in \pa{B}
+\[ \eqn{ Create Acyclic }{
+ \pendsof{L}{\pqy} = \{ \}
}\]
\subsection{No Replay}
-If $\patchof{L} = \pa{L}$, trivial by Base Acyclic for $L$.
-
-If $\patchof{L} = \bot$, consider some $D \isin B$. $D \neq B$.
-Thus $D \isin L$. So by No Replay of $D$ in $L$, $D \le L$.
-Thus $D \le B$.
+Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
\subsection{Base Acyclic}
-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.
-
-$\qed$
+Consider some $D \isin B$. If $D = B$, $D \in \pqn$.
+If $D \neq B$, $D \isin L$, so by No Replay $D \le L$
+and by Create Acyclic
+$D \not\in \pqy$. $\qed$
\subsection{Coherence and Patch Inclusion}
-Consider some $D \in \p$.
-$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$.
+Consider some $D \in \py$.
+$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$
+and $D \le B \equiv D \le L$.
Thus $L \haspatch \p \implies B \haspatch P$
and $L \nothaspatch \p \implies B \nothaspatch P$.
\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 $D \isin B \equiv D \le B$. $\qed$
+Simple Foreign Inclusion applies. $\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
\section{Create Tip}
-xxx tbd
+Given a Topbloke base $B$ for a patch $\pq$,
+create a tip branch initial commit B.
+\gathbegin
+ C \hasparents \{ B \}
+\gathnext
+ \patchof{B} = \pqy
+\gathnext
+ D \isin C \equiv D \isin B \lor D = C
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{B} = \pqn
+}\]
+\[ \eqn{ No Sneak }{
+ \pendsof{B}{\pqy} = \{ \}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. $\qed$
+
+\subsection{Tip Contents}
+
+Consider some arbitrary commit $D$. If $D = C$, trivially satisfied.
+
+If $D \neq C$, $D \isin C \equiv D \isin B$,
+which by Unique Base, above, $ \equiv D \isin \baseof{B}$.
+By Base Acyclic of $B$, $D \isin B \implies D \not\in \pqy$.
+
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+Not applicable.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+ \p = \pq \lor B \haspatch \p : & C \haspatch \p \\
+ \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
-\section{Anticommit}
+\proofstarts
+~ Consider some $D \in \py$.
+
+\subsubsection{For $\p = \pq$:}
+
+By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
+By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \haspatch \pq$.
+
+\subsubsection{For $\p \neq \pq$:}
+
+$D \neq C$. So $D \isin C \equiv D \isin B$,
+and $D \le C \equiv D \le B$.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Simple Foreign Inclusion applies. $\qed$
-Given $L$ and $\pr$ as represented by $R^+, R^-$.
-Construct $C$ which has $\pr$ removed.
+\subsection{Foreign Contents}
+
+Not applicable.
+
+\section{Dependency Removal}
+
+Given $L$ which contains $\pr$ as represented by $R^+, R^-$.
+Construct $C$ which has $\pr$ removed by applying a single
+commit which is the anticommit of $\pr$.
Used for removing a branch dependency.
\gathbegin
C \hasparents \{ L \}
R^+ \in \pry \land R^- = \baseof{R^+}
}\]
\[ \eqn{ Into Base }{
- L \in \pn
+ L \in \pqn
}\]
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
L \haspatch \pry
}\]
-\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 $\merge$,
+$D \isin C \implies D \isin L \lor D \isin R^- \lor D = C$.
+So, by Ordering of Ingredients,
+Ingredients Prevent Replay applies. $\qed$
\subsection{Desired Contents}
\subsubsection{For $D \neq C, D \not\le L$:}
-By No Replay $D \not\isin L$. Also $D \not\le R^-$ hence
+By No Replay for $L$, $D \not\isin L$.
+Also, by Ordering of Ingredients, $D \not\le R^-$ hence
$D \not\isin R^-$. Thus $D \not\isin C$. OK.
\subsubsection{For $D \neq C, D \le L, D \in \pry$:}
By Currently Included, $D \isin L$.
-By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by
-by Unique Tip, $D \le R^+ \equiv D \le L$.
+By Tip Self Inpatch for $R^+$, $D \isin R^+ \equiv D \le R^+$, but by
+by Unique Tip, $D \le R^+ \equiv D \le L$.
So $D \isin R^+$.
-By Base Acyclic, $D \not\isin R^-$.
+By Base Acyclic for $R^-$, $D \not\isin R^-$.
Apply $\merge$: $D \not\isin C$. OK.
\subsection{Unique Base}
-Into Base means that $C \in \pn$, so Unique Base is not
+Into Base means that $C \in \pqn$, so Unique Base is not
applicable. $\qed$
\subsection{Tip Contents}
\subsection{Base Acyclic}
-By Base Acyclic for $L$, $D \isin L \implies D \not\in \py$.
-And by Into Base $C \not\in \py$.
+By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \pqy$.
+And by Into Base $C \not\in \pqy$.
Now from Desired Contents, above, $D \isin C
\implies D \isin L \lor D = C$, which thus
-$\implies D \not\in \py$. $\qed$.
+$\implies D \not\in \pqy$. $\qed$.
\subsection{Coherence and Patch Inclusion}
Not applicable.
+\section{Dependency Insertion}
+
+Given $L$ construct $C$ which additionally
+contains $\pr$ as represented by $R^+$ and $R^-$.
+This may even be used for reintroducing a previous-removed branch
+dependency.
+\gathbegin
+ C \hasparents \{ L, R^+ \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \mergeof{C}{L}{R^-}{R^+}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ R^- = \baseof{R^+}
+}\]
+\[ \eqn{ Into Base }{
+ L \in \pqn
+}\]
+\[ \eqn{ Currently Excluded }{
+ L \nothaspatch \pr
+}\]
+\[ \eqn{ Inserted's Ends }{
+ E \in \pendsof{L}{\pry} \implies E \le R^+
+}\]
+\[ \eqn{ Others' Ends }{
+ \bigforall_{\p \patchisin \L}
+ E \in \pendsof{R^+}{\py} \implies E \le L
+}\]
+\[ \eqn{ Insertion Acyclic }{
+ R^+ \nothaspatch \pq
+}\]
+
+\subsection{No Replay}
+
+By $\merge$,
+$D \isin C \implies D \isin L \lor D \isin R^+ \lor D = C$.
+So Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Not applicable.
+
+\subsection{Tip Contents}
+
+Not applicable.
+
+\subsection{Base Acyclic}
+
+Consider some $D \isin C$. We will show that $D \not\in \pqy$.
+By $\merge$, $D \isin L \lor D \isin R^+ \lor D = C$.
+
+For $D \isin L$, Base Acyclic for L suffices. For $D \isin R^+$,
+Insertion Acyclic suffices. For $D = C$, trivial. $\qed$.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+ \p = \pr \lor L \haspatch \p : & C \haspatch \p \\
+ \p \neq \pr \land L \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+\proofstarts
+~ Consider some $D \in \py$.
+$D \neq C$ so $D \le C \equiv D \le L \lor D \le R^+$.
+
+\subsubsection{For $\p = \pr$:}
+
+$D \not\isin L$ by Currently Excluded.
+$D \not\isin R^-$ by Base Acyclic.
+So by $\merge$, $D \isin C \equiv D \isin R^+$,
+which by Tip Self Inpatch of $R^+$, $\equiv D \le R^+$.
+
+And by definition of $\pancs$,
+$D \le L \equiv D \in \pancsof{L}{R^+}$.
+Applying Transitive Ancestors to Inserted's Ends gives
+$A \in \pancsof{L}{R^+} \implies A \le R^+$.
+So $D \le L \implies D \le R^+$.
+Thus $D \le C \equiv D \le R^+$.
+
+So $D \isin C \equiv D \le C$, i.e. $C \haspatch \pr$.
+OK.
+
+\subsubsection{For $\p \neq \pr$:}
+
+By Exclusive Tip Contents for $R^+$ ($D \not\in \pry$ case)
+$D \isin R^+ \equiv D \isin R^-$.
+So by $\merge$, $D \isin C \equiv D \isin L$.
+
+If $L \nothaspatch \p$, $D \not\isin L$ so $C \nothaspatch \p$. OK.
+
+If $L \haspatch \p$, Others' Ends applies; by Transitive
+Ancestors, $A \in \pancsof{R^+}{\py} \implies A \le L$.
+So $D \le R^+ \implies D \le L$,
+since $D \le R^+ \equiv D \in \pancsof{R^+}{\py}$.
+Thus $D \le C \equiv D \le L$.
+And by $\haspatch$, $D \le L \equiv D \isin L$ so
+$D \isin C \equiv D \le C$. Thus $C \haspatch \p$.
+OK.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+
+By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
+So by $\merge$, $D \isin C \equiv D \isin L$.
+
+xxx up to here, need new condition
+
+$D \neq C$.
+
+
+
\section{Merge}
Merge commits $L$ and $R$ using merge base $M$:
Merge Ends condition applies.
So a plain git merge of non-Topbloke branches meets the conditions and
-is therefore consistent with our scheme.
+is therefore consistent with our model.
\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}
putative ancestor $A \in \pn$ and see whether $A \le C$.
By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
-But $C \in py$ and $A \in \pn$ so $A \neq C$.
+But $C \in py$ and $A \in \pn$ so $A \neq C$.
Thus $A \le C \equiv A \le L \lor A \le R$.
By Unique Base of L and Transitive Ancestors,
But by Tip Merge condition on $\baseof{R}$,
$A \le \baseof{L} \implies A \le \baseof{R}$, so
$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.
That is, $\baseof{C} = \baseof{R}$.
\subsubsection{For $R \in \pn$:}
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 $A \le C \equiv A \le R$.
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.
+Thus $A \le C \equiv A \le R$.
That is, $\baseof{C} = R$.
$\qed$
\subsection{Coherence and Patch Inclusion}
Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
-This involves considering $D \in \py$.
+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 $D \neq C$.
+\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch for $L$). So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
- \equiv D \isin L \lor D \isin R$.
+ \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 $D \le X$
+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 $D \not\le C$.
+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 $D \not\isin M$.
+$D \not\le X$ 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 $C \haspatch \p$.
\proofstarts
-One of the Merge Ends conditions 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
(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,
+Self Inpatch for $L$, $L \haspatch \p$, 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 $D \le C$.
+$D \le Y$ 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
+$D \in \pancsof{X}{\py}$, 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.
\subsection{Tip Contents}
-We need worry only about $C \in \py$.
+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.
~ian / topbloke-formulae.git / blobdiff