\section{Notation}
-xxx make all conditions be in conditions, not in (or also in) intro
-
\begin{basedescript}{
\desclabelwidth{5em}
\desclabelstyle{\nextlinelabel}
\[\eqn{Foreign Inclusion:}{
\bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
}\]
+\[\eqn{Foreign Contents:}{
+ \bigforall_{C \text{ s.t. } \patchof{C} = \bot}
+ D \le C \implies \patchof{D} = \bot
+}\]
\section{Some lemmas}
}\]
xxx proof tbd
-\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.
+\[ \eqn{Ingredients Prevent Replay:}{
+ \left[
+ {C \hasparents \set A} \land
+ \\
+ \left(
+ D \isin C \implies
+ D = C \lor
+ \Largeexists_{A \in \set A} D \isin A
+ \right)
+ \right] \implies \left[
+ D \isin C \implies D \le C
+ \right]
+}\]
+\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 $D \le C$.
+}
-$\qed$
+\[ \eqn{Totally Foreign Contents:}{
+ \bigforall_{C \hasparents \set A}
+ \left[
+ \patchof{C} = \bot \land
+ \bigforall_{A \in \set A} \patchof{A} = \bot
+ \right]
+ \implies
+ \left[
+ D \le C
+ \implies
+ \patchof{D} = \bot
+ \right]
+}\]
+\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}
-xxx need to compute annotation for every commit type
-
We annotate each Topbloke commit $C$ with:
\gathbegin
\patchof{C}
\bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
\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.
+
+$C \haspatch \pa{Q}$ or $\nothaspatch \pa{Q}$ is represented as the
+set $\{ \pa{Q} | C \haspatch \pa{Q} \}$. Whether $C \haspatch \pa{Q}$
+is in stated
+(in terms of $I \haspatch \pa{Q}$ or $I \nothaspatch \pa{Q}$
+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,
+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}
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 $A, C \in \py$ then by Calculation of Ends for
+$C, \py, C \not\in \py$:
+$\pendsof{C}{\pn} = \pendsof{A}{\pn}$ so
+$\baseof{C} = \baseof{A}$. $\qed$
\subsection{Tip Contents}
We need to consider only $A, C \in \py$. From Tip Contents for $A$:
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
-$A$, $D \isin C \implies D \not\in \py$. $\qed$
+$A$, $D \isin C \implies D \not\in \py$.
+
+$\qed$
\subsection{Coherence and patch inclusion}
If $D = C$, trivial. For $D \neq C$:
$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
+\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} = \pan{Q}
+\gathnext
+ D \isin B \equiv D \isin L \lor D = B
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{L} = \pa{L} \lor \patchof{L} = \bot
+}\]
+\[ \eqn{ Non-recursion }{
+ L \not\haspatch \pa{Q}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Not applicable.
+
+\subsection{Tip Contents}
+
+Not applicable.
+
+\subsection{Base Acyclic}
+
+Consider some $D \isin B$. If $D = B$, $D \in \pan{Q}$.
+If $D \neq B$, $D \isin L$, and by Non-recursion
+$D \not\in \pay{Q}$. $\qed$
+
+\subsection{Coherence and Patch Inclusion}
+
+Consider some $D \in \p$.
+$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$.
+
+Thus $L \haspatch \p \implies B \haspatch P$
+and $L \nothaspatch \p \implies B \nothaspatch P$.
+
+$\qed$.
+
+\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 $D \isin B \equiv D \le B$. $\qed$
+
+\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} = \pay{Q}
+\gathnext
+ D \isin C \equiv D \isin B \lor D = C
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{B} = \pan{Q}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Trivially, $\pendsof{C}{\pan{Q}} = \{B\}$ so $\baseof{C} = B$.
+
+\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 $B$, $D \isin B \implies D \not\in \pay{Q}$.
+So $D \isin C \equiv D \isin \baseof{B}$.
+
+$\qed$
+
+xxx up to here
+
\section{Anticommit}
-Given $L, R^+, R^-$ where
-$R^+ \in \pry, R^- = \baseof{R^+}$.
+Given $L$ and $\pr$ as represented by $R^+, R^-$.
Construct $C$ which has $\pr$ removed.
Used for removing a branch dependency.
\gathbegin
\subsection{Conditions}
+\[ \eqn{ Ingredients }{
+R^+ \in \pry \land R^- = \baseof{R^+}
+}\]
\[ \eqn{ Into Base }{
L \in \pn
}\]
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 $R^- \le C$.
+$\qed$
-(Note that the merge base $R^+ \not\le R^-$, i.e. the merge base is
-later than one of the branches to be merged.)
+(Note that $R^+ \not\le R^-$, i.e. the merge base
+is a descendant, not an ancestor, of the 2nd parent.)
\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 Ordering of Ingredients,
+Ingredients Prevent Replay applies. $\qed$
\subsection{Desired Contents}
Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
so $L \haspatch \p \implies C \haspatch \p$.
-\section{Foreign Inclusion}
+$\qed$
+
+\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 $D \isin C \equiv D \le C$. $\qed$
+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
We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\subsection{Conditions}
-
+\[ \eqn{ Ingredients }{
+ M \le L, M \le R
+}\]
\[ \eqn{ Tip Merge }{
L \in \py \implies
\begin{cases}
\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
+\[ \eqn{ Foreign Merges }{
+ \patchof{L} = \bot \equiv \patchof{R} = \bot
+}\]
\subsection{Non-Topbloke merges}
-We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$.
+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.
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. Good.
+Merge Ends condition applies.
+
+So a plain git merge of non-Topbloke branches meets the conditions and
+is therefore consistent with our scheme.
\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}
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 $D \not\isin C$. $\qed$
+R$. And $D \neq C$. So $D \not\isin C$.
+
+$\qed$
\subsection{Tip Contents}
$\qed$
+\subsection{Foreign Contents}
+
+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}
~ian / topbloke-formulae.git / blobdiff