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 $\bot$.
$\displaystyle \bigforall_{D \in \py} D \not\isin C $.
~ 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
}
\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 )
}\]
\subsection{Tip Self Inpatch}
+Given Exclusive Tip Contents and Base Acyclic for $C$,
$$
\bigforall_{C \in \py} C \haspatch \p
$$
\subsection{Exact Ancestors}
$$
\bigforall_{ C \hasparents \set{R} }
+ \left[
D \le C \equiv
( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
\lor D = C
+ \right]
$$
\proof{ ~ Trivial.}
by the LHS. And $A \le A''$.
}
-\subsection{Calculation Of Ends}
+\subsection{Calculation of Ends}
$$
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
\left[
{C \hasparents \set A} \land
\\
+ \bigforall_{D}
\left(
- D \isin C \implies
+ D \isin C \implies
D = C \lor
\Largeexists_{A \in \set A} D \isin A
\right)
- \right] \implies \left[
+ \right] \implies \left[ \bigforall_{D}
D \isin C \implies D \le C
\right]
$$
\subsection{Totally Foreign Contents}
$$
- \bigforall_{C \hasparents \set A}
\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
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 $I$),
+for the ingredients $I$)
in the proof of Coherence for each kind of commit.
$\pendsof{C}{\pq^+}$ is computed, for all Topbloke-generated commits,
Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
-If $L, C \in \py$ then by Calculation of Ends for
-$C, \py, C \not\in \py$:
+If $L, C \in \py$ then by Calculation of Ends,
$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ so
$\baseof{C} = \baseof{L}$. $\qed$
\[ D \isin C \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L )
\lor D = C \]
Since $D = C \implies D \in \py$,
-and substituting in $\baseof{C}$, this gives:
+and substituting in $\baseof{C}$, from Unique Base above, this gives:
\[ D \isin C \equiv D \isin \baseof{C} \lor
(D \in \py \land D \le L) \lor
(D = C \land D \in \py) \]
\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
\subsection{Conditions}
-\[ \eqn{ Ingredients }{
- \patchof{L} = \pa{L} \lor \patchof{L} = \bot
-}\]
\[ \eqn{ Create Acyclic }{
- L \not\haspatch \pq
+ \pendsof{L}{\pqy} = \{ \}
}\]
\subsection{No Replay}
\subsection{Base Acyclic}
Consider some $D \isin B$. If $D = B$, $D \in \pqn$.
-If $D \neq B$, $D \isin L$, and by Create Acyclic
+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 $D \in \p$.
+Consider some $D \in \py$.
$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$
and $D \le B \equiv D \le L$.
\section{Create Tip}
-Given a Topbloke base $B$, create a tip branch initial commit B.
+Given a Topbloke base $B$ for a patch $\pq$,
+create a tip branch initial commit B.
\gathbegin
C \hasparents \{ B \}
\gathnext
Consider some arbitrary commit $D$. If $D = C$, trivially satisfied.
-If $D \neq C$, $D \isin C \equiv D \isin B$.
+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 $B$, $D \isin B \implies D \not\in \pqy$.
-So $D \isin C \equiv D \isin \baseof{B}$.
+
$\qed$
\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 $C \haspatch \pq$.
+By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \haspatch \pq$.
\subsubsection{For $\p \neq \pq$:}
Not applicable.
-\section{Anticommit}
+\section{Dependency Removal}
-Given $L$ and $\pr$ as represented by $R^+, R^-$.
-Construct $C$ which has $\pr$ removed.
+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^+ \}
\subsection{No Replay}
-By definition of $\merge$,
+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. $\qed$
\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 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}
\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 $L \haspatch \p, R \haspatch \p$:}
(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.