\documentclass[a4paper,leqno]{strayman}
+\errorcontextlines=50
\let\numberwithin=\notdef
\usepackage{amsmath}
\usepackage{mathabx}
\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{\nle}{\nleqslant}
\newcommand{\has}{\sqsupseteq}
\newcommand{\isin}{\sqsubseteq}
\newif\ifhidehack\hidehackfalse
\DeclareRobustCommand\hidefromedef[2]{%
\hidehacktrue\ifhidehack#1\else#2\fi\hidehackfalse}
- \newcommand{\pa}[1]{\hidefromedef{\varmathbb #1}{#1}}
+ \newcommand{\pa}[1]{\hidefromedef{\varmathbb{#1}}{#1}}
-\newcommand{\set}[1]{\mathbb #1}
+\newcommand{\set}[1]{\mathbb{#1}}
\newcommand{\pay}[1]{\pa{#1}^+}
\newcommand{\pan}[1]{\pa{#1}^-}
\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}}
+
%\newcommand{\hasparents}{\underaccent{1}{>}}
%\newcommand{\hasparents}{{%
% \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}}
\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
\newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
\newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
-\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
-\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
+\newcommand{\merge}{{\mathcal M}}
+\newcommand{\mergeof}[4]{\merge(#1,#2,#3,#4)}
+%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
+\newcommand{\patch}{{\mathcal P}}
+\newcommand{\base}{{\mathcal B}}
+
+\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}{%
\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*{}}
\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 $\pn$.
-All these sets are distinct. Hence:
+None of these sets overlap. 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 $\p$; if a Topbloke
+Non-Topbloke commits 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 Topbloke patch itself, we hope that git's merge algorithm will
-DTRT or that the user will no longer care about the Topbloke patch.
+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.
+
+\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
+The contents of a git merge result:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ (D \isin L \land D \isin R) \lor D = C : & \true \\
+ (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
+ \text{otherwise} : & D \not\isin M
+ \end{cases}
+$
\end{basedescript}
\newpage
\[\eqn{Foreign Inclusion:}{
\bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
}\]
+\[\eqn{Foreign Contents:}{
+ \bigforall_{C \text{ s.t. } \patchof{C} = \bot}
+ D \le C \implies \patchof{D} = \bot
+}\]
\section{Some lemmas}
-\[ \eqn{Exclusive Tip Contents:}{
- \bigforall_{C \in \py}
+\subsection{Alternative (overlapping) formulations of $\mergeof{C}{L}{M}{R}$}
+$$
+ D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin L \equiv D \isin M : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin M : & D = C \lor D \isin L \\
+ \text{as above with L and R exchanged}
+ \end{cases}
+$$
+\proof{ ~ Truth table (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 $R$.
+}
+
+\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{
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{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
\bigforall_{C \in \py} D \isin C \equiv
\begin{cases}
D \in \py : & D \le C \\
\end{cases}
}\]
-\[ \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
-}\]
+$$
+\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,
commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
by the LHS. And $A \le A''$.
}
-\[ \eqn{Calculation Of Ends:}{
+
+\subsection{Calculation Of Ends}
+$$
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
- \Bigl\{ E \Big|
- \Bigl[ \Largeexists_{A \in \set A}
+ \begin{cases}
+ C \in \p : & \{ C \}
+ \\
+ C \not\in \p : & \displaystyle
+ \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]
- \Bigr\}
-}\]
-XXX proof TBD.
+ \Bigl[ \Largenexists_{B \in \set A, F \in \pendsof{B}{\p}}
+ E \neq F \land E \le F \Bigr]
+ \right\}
+ \end{cases}
+$$
+\proof{
+Trivial for $C \in \set P$. For $C \not\in \set P$,
+$\pancsof{C}{\set P} = \bigcup_{A \in \set A} \pancsof{A}{\set P}$.
+So $\pendsof{C}{\set P} \subset \bigcup_{E in \set E} \pendsof{E}{\set P}$.
+Consider some $E \in \pendsof{A}{\set P}$. If $\exists_{B,F}$ as
+specified, then either $F$ is going to be in our result and
+disqualifies $E$, or there is some other $F'$ (or, eventually,
+an $F''$) which disqualifies $F$.
+Otherwise, $E$ meets all the conditions for $\pends$.
+}
+
+\subsection{Ingredients Prevent Replay}
+$$
+ \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]
+$$
+\proof{
+ Trivial for $D = C$. Consider some $D \neq C$, $D \isin C$.
+ By the preconditions, there is some $A$ s.t. $D \in \set A$
+ and $D \isin A$. By No Replay for $A$, $D \le A$. And
+ $A \le C$ so $D \le C$.
+}
+
+\subsection{Simple Foreign Inclusion}
+$$
+ \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]
+$$
+\proof{
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+If $D = C$, trivially true. For $D \neq C$,
+by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
+
+\subsection{Totally Foreign Contents}
+$$
+ \bigforall_{C \hasparents \set A}
+ \left[
+ \patchof{C} = \bot \land
+ \bigforall_{A \in \set A} \patchof{A} = \bot
+ \right]
+ \implies
+ \left[
+ 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 $A$, $\patchof{D} = \bot$.
+}
\section{Commit annotation}
\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
+in the summary in the section for that kind of commit.
+
+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.
+
+$C \haspatch \pq$ or $\nothaspatch \pq$ is represented as the
+set $\{ \pq | C \haspatch \pq \}$. Whether $C \haspatch \pq$
+is in stated
+(in terms of $I \haspatch \pq$ or $I \nothaspatch \pq$
+for the ingredients $I$),
+in the proof of Coherence for each kind of commit.
+
+$\pendsof{C}{\pq^+}$ 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 $\pends$
-would have to be updated. The annotation is not needed because
-$\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
+would have to be updated. The annotation is not needed in that case
+because $\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
\section{Simple commit}
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 $\baseof{C} = \baseof{A}$. $\qed$
+If $L, C \in \py$ then by Calculation of Ends for
+$C, \py, C \not\in \py$:
+$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ 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 $D = C \implies D \in \py$,
+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
-$A$, $D \isin C \implies D \not\in \py$. $\qed$
+For $D \neq C$: $D \isin C \equiv D \isin L$, so by Base Acyclic for
+$L$, $D \isin C \implies D \not\in \py$.
+
+$\qed$
\subsection{Coherence and patch inclusion}
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:}
+
+Simple Foreign Inclusion applies. $\qed$
+
+\subsection{Foreign Contents:}
+
+Only relevant if $\patchof{C} = \bot$, and in that case Totally
+Foreign Contents applies. $\qed$
+
+\section{Create Base}
+
+Given $L$, create a Topbloke base branch initial commit $B$.
+\gathbegin
+ B \hasparents \{ L \}
+\gathnext
+ \patchof{B} = \pqn
+\gathnext
+ D \isin B \equiv D \isin L \lor D = B
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Create Acyclic }{
+ \pendsof{L}{\pqy} = \{ \}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Not applicable.
-If $D = C$, trivial. For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
+\subsection{Tip Contents}
+
+Not applicable.
+
+\subsection{Base Acyclic}
+
+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 $D \isin B \equiv D \isin L$
+and $D \le B \equiv D \le L$.
+
+Thus $L \haspatch \p \implies B \haspatch P$
+and $L \nothaspatch \p \implies B \nothaspatch P$.
+
+$\qed$.
+
+\subsection{Foreign Inclusion}
+
+Simple Foreign Inclusion applies. $\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
+
+\section{Create Tip}
+
+Given a Topbloke base $B$, 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 $D \neq C$, $D \isin C \equiv D \isin B$.
+By Base Acyclic of $B$, $D \isin B \implies D \not\in \pqy$.
+So $D \isin C \equiv D \isin \baseof{B}$.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+Not applicable.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+ \p = \pq \lor B \haspatch \p : & C \haspatch \p \\
+ \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+
+\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, $D \le B \equiv D = C$. 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. $\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
+
+\section{Anticommit}
+
+Given $L$ and $\pr$ as represented by $R^+, R^-$.
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \mergeof{C}{L}{R^+}{R^-}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+R^+ \in \pry \land R^- = \baseof{R^+}
+}\]
+\[ \eqn{ Into Base }{
+ L \in \pn
+}\]
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+
+\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 $R^- \le C$.
+$\qed$
+
+(Note that $R^+ \not\le R^-$, i.e. the merge base
+is a descendant, not an ancestor, of the 2nd parent.)
+
+\subsection{No Replay}
+
+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}
+
+\[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \]
+\proofstarts
+
+\subsubsection{For $D = C$:}
+
+Trivially $D \isin C$. OK.
+
+\subsubsection{For $D \neq C, D \not\le L$:}
+
+By No Replay $D \not\isin L$. Also $D \not\le R^-$ 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, $D \le R^+ \equiv D \le L$.
+So $D \isin R^+$.
+
+By Base Acyclic, $D \not\isin R^-$.
+
+Apply $\merge$: $D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C, D \le L, D \notin \pry$:}
+
+By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
+
+Apply $\merge$: $D \isin C \equiv D \isin L$. OK.
+
+$\qed$
+
+\subsection{Unique Base}
+
+Into Base means that $C \in \pn$, so Unique Base is not
+applicable. $\qed$
+
+\subsection{Tip Contents}
+
+Again, not applicable. $\qed$
+
+\subsection{Base Acyclic}
+
+By Base Acyclic for $L$, $D \isin L \implies D \not\in \py$.
+And by Into Base $C \not\in \py$.
+Now from Desired Contents, above, $D \isin C
+\implies D \isin L \lor D = C$, which thus
+$\implies D \not\in \py$. $\qed$.
+
+\subsection{Coherence and Patch Inclusion}
+
+Need to consider some $D \in \py$. By Into Base, $D \neq C$.
+
+\subsubsection{For $\p = \pr$:}
+By Desired Contents, above, $D \not\isin C$.
+So $C \nothaspatch \pr$.
+
+\subsubsection{For $\p \neq \pr$:}
+By Desired Contents, $D \isin C \equiv D \isin L$
+(since $D \in \py$ so $D \not\in \pry$).
+
+If $L \nothaspatch \p$, $D \not\isin L$ so $D \not\isin C$.
+So $L \nothaspatch \p \implies C \nothaspatch \p$.
+
+Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
+so $L \haspatch \p \implies C \haspatch \p$.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$.
+So by Desired Contents $D \isin C \equiv D \isin L$.
+By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+
+And $D \le C \equiv D \le L$.
+Thus $D \isin C \equiv D \le C$.
+
+$\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
\section{Merge}
-Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
+Merge commits $L$ and $R$ using merge base $M$:
\gathbegin
C \hasparents \{ L, R \}
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- D \isin C \equiv
- \begin{cases}
- (D \isin L \land D \isin R) \lor D = C : & \true \\
- (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
- \text{otherwise} : & D \not\isin M
- \end{cases}
+ \mergeof{C}{L}{M}{R}
\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\subsection{Conditions}
-
+\[ \eqn{ Ingredients }{
+ M \le L, M \le R
+}\]
\[ \eqn{ Tip Merge }{
L \in \py \implies
\begin{cases}
R \in \py : & \baseof{R} \ge \baseof{L}
\land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
- R \in \pn : & R \ge \baseof{L}
- \land M = \baseof{L} \\
+ R \in \pn : & M = \baseof{L} \\
\text{otherwise} : & \false
\end{cases}
}\]
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
+\[ \eqn{ Removal Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \haspatch \p
+ \implies
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
+}\]
+\[ \eqn{ Foreign Merges }{
+ \patchof{L} = \bot \equiv \patchof{R} = \bot
+}\]
-\subsection{No Replay}
-
-\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.
+\subsection{Non-Topbloke merges}
-\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
-$D \not\isin C$. OK.
+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.
-\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
-$D \not\isin C$. OK.
+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.
-\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.
+So a plain git merge of non-Topbloke branches meets the conditions and
+is therefore consistent with our model.
-\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \isin M$:}
-$D \not\isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
-R$ so $D \le C$. OK.
+\subsection{No Replay}
-$\qed$
+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}
and calculate $\pendsof{C}{\pn}$. So we will consider some
putative ancestor $A \in \pn$ and see whether $A \le C$.
-$A \le C \equiv A \le L \lor A \le R \lor A = C$.
-But $C \in py$ and $A \in \pn$ so $A \neq C$.
-Thus $A \le L \lor A \le R$.
+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 $A \neq C$.
+Thus $A \le C \equiv A \le L \lor A \le R$.
By Unique Base of L and Transitive Ancestors,
$A \le L \equiv A \le \baseof{L}$.
But by Tip Merge condition on $\baseof{R}$,
$A \le \baseof{L} \implies A \le \baseof{R}$, so
-$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} =
-\baseof{R}$.
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+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 $A \le C \equiv A \le R$.
+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 $D \in \py$.
+
+\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 $D \isin Y \equiv D \le Y$.)
-UP TO HERE
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
-By Tip Merge, $A \le $
+For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
+ \equiv D \isin L \lor D \isin R$.
+(Likewise $D \le C \equiv D \le X \lor D \le Y$.)
-Let $S =
- \begin{cases}
- R \in \py : & \baseof{R} \\
- R \in \pn : & R
- \end{cases}$.
-Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
-
-Consider some $A \in \pn$. If $A \le S$ then $A \le C$.
-If $A \not\le S$ then
+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 $C \haspatch \p$.
-Let $A \in \pends{C}{\pn}$.
-Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
-\pendsof{R,\pn}$.
+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 $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$M \haspatch \p \implies C \nothaspatch \p$.
+$M \nothaspatch \p \implies C \haspatch \p$.
+
+\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.
+
+$\qed$
+
+\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 $D \not\in \py, R \in \py$:}
+
+$D \neq C$.
+
+By Tip Contents
+$D \isin L \equiv D \isin \baseof{L}$ and
+$D \isin R \equiv D \isin \baseof{R}$.
+
+If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
+Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
+$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
+$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
+
+So $D \isin M \equiv D \isin L$ and by $\merge$,
+$D \isin C \equiv D \isin R$. But from Unique Base,
+$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$. OK.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+By Foreign Inclusion of $L, M, R$:
+$D \isin L \equiv D \le L$;
+$D \isin M \equiv D \le M$;
+$D \isin R \equiv D \le R$.
+
+\subsubsection{For $D = C$:}
+
+$D \isin C$ and $D \le C$. OK.
+
+\subsubsection{For $D \neq C, D \isin M$:}
+
+Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
+R$. So by $\merge$, $D \isin C$. And $D \le C$. OK.
+
+\subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
+
+By $\merge$, $D \isin C$.
+And $D \isin X$ means $D \le X$ so $D \le C$.
+OK.
+
+\subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
+
+By $\merge$, $D \not\isin C$.
+And $D \not\le L, D \not\le R$ so $D \not\le C$.
+OK
+
+$\qed$
-%$\pends{C,
+\subsection{Foreign Contents}
-%%\subsubsection{For $R \in \py$:}
+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}