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strategy: calculate/use \allsrcs
[topbloke-formulae.git]
/
merge.tex
diff --git
a/merge.tex
b/merge.tex
index 71ad81fbc77d71247f48bbe786242696aa6073d9..b43d30fea899e45b07d47c1aae08d67eb2b7ee2a 100644
(file)
--- a/
merge.tex
+++ b/
merge.tex
@@
-10,8
+10,9
@@
Merge commits $L$ and $R$ using merge base $M$:
\end{gather}
We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\end{gather}
We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
-This can also be used for dependency re-insertion, by setting
-$L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
+This can also be used for dependency re-insertion, by setting $L \in
+\pn$, $R \in \pry$, $M = \baseof{R}$, provided that the Conditions are
+satisfied; in particular, provided that $L \ge \baseof{R}$.
\subsection{Conditions}
\[ \eqn{ Ingredients }{
\subsection{Conditions}
\[ \eqn{ Ingredients }{
@@
-46,20
+47,28
@@
$L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
+\[ \eqn{ Suitable Tips }{
+ \bigforall_{\p \neq \patchof{L}, \; C \haspatch \p}
+ \bigexists_T
+ \pendsof{J}{\py} = \{ T \}
+ \land
+ \forall_{E \in \pendsof{K}{\py}} T \ge E
+ , \text{where} \{J,K\} = \{L,R\}
+}\]
\[ \eqn{ Foreign Merges }{
\[ \eqn{ Foreign Merges }{
- \
patchof{L} = \bot \implies \patchof{R} = \bot
+ \
isforeign{L} \implies \isforeign{R}
}\]
\subsection{Non-Topbloke merges}
}\]
\subsection{Non-Topbloke merges}
-We require both $\
patchof{L} = \bot$ and $\patchof{R} = \bot
$
+We require both $\
isforeign{L}$ and $\isforeign{R}
$
(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.
(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.
-By Foreign Contents of $L$, $\
patchof{M} = \bot
$ as well.
+By Foreign Contents of $L$, $\
isforeign{M}
$ as well.
So by Foreign Contents for any $A \in \{L,M,R\}$,
$\forall_{\p, D \in \py} D \not\le A$
so $\pendsof{A}{\py} = \{ \}$ and the RHS of both Merge Ends
So by Foreign Contents for any $A \in \{L,M,R\}$,
$\forall_{\p, D \in \py} D \not\le A$
so $\pendsof{A}{\py} = \{ \}$ and the RHS of both Merge Ends
@@
-111,14
+120,23
@@
$\qed$
\subsection{Coherence and Patch Inclusion}
\subsection{Coherence and Patch Inclusion}
-Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
-This involves considering $D \in \py$.
+$$
+\begin{cases}
+ L \nothaspatch \p \land R \nothaspatch \p : & C \nothaspatch \p \\
+ L \haspatch \p \land R \haspatch \p : & C \haspatch \p \\
+ \text{otherwise} \land M \haspatch \p : & C \nothaspatch \p \\
+ \text{otherwise} \land M \nothaspatch \p : & C \haspatch \p
+\end{cases}
+$$
+\proofstarts
+~ Consider $D \in \py$.
\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
\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 $
\neg[ L \nothaspatch \p ]
$ by Tip Own Contents for $L$).
+\in \py$ ie $
L \haspatch \p
$ by Tip Own Contents for $L$).
So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+OK.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
@@
-150,25
+168,21
@@
and this $F \le C$ so indeed $C \haspatch \p$.
\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
\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
if $M \haspatch \p$, $D \not\isin C$,
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
if $M \haspatch \p$, $D \not\isin C$,
-whereas if $M \nothaspatch \p$, $D \isin C \equiv
\land
D \le C$.
+whereas if $M \nothaspatch \p$, $D \isin C \equiv D \le C$.
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.
Consider $D = C$: Thus $C \in \py, L \in \py$.
And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this
$F \le C$ so this suffices.
Consider $D = C$: Thus $C \in \py, L \in \py$.
-By Tip Own Contents, $
\neg[ L \nothaspatch \p ]
$ so $L \neq X$,
+By Tip Own Contents, $
L \haspatch \p
$ so $L \neq X$,
therefore we must have $L=Y$, $R=X$.
therefore we must have $L=Y$, $R=X$.
-By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
+Conversely $R \not\in \py$
+so by Tip Merge $M = \baseof{L}$. Thus $M \in \pn$ so
by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
and $D \le C$. OK.
by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
and $D \le C$. OK.
@@
-253,9
+267,20
@@
Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
$\qed$
$\qed$
+\subsection{Unique Tips}
+
+For $L \in \py$, trivially $\pendsof{C}{\py} = C$ so $T = C$ is
+suitable.
+
+For $L \not\in \py$, $\pancsof{C}{\py} = \pancsof{L}{\py} \cup
+\pancsof{R}{\py}$. So $T$ from Suitable Tips is a suitable $T$ for
+Unique Tips.
+
+$\qed$
+
\subsection{Foreign Inclusion}
\subsection{Foreign Inclusion}
-Consider some $D
$ s.t. $\patchof{D} = \bot
$.
+Consider some $D
\in \foreign
$.
By Foreign Inclusion of $L, M, R$:
$D \isin L \equiv D \le L$;
$D \isin M \equiv D \le M$;
By Foreign Inclusion of $L, M, R$:
$D \isin L \equiv D \le L$;
$D \isin M \equiv D \le M$;
@@
-286,6
+311,6
@@
$\qed$
\subsection{Foreign Contents}
\subsection{Foreign Contents}
-Only relevant if $\
patchof{L} = \bot
$, in which case
-$\
patchof{C} = \bot$ and by Foreign Merges $\patchof{R} = \bot
$,
+Only relevant if $\
isforeign{L}
$, in which case
+$\
isforeign{C}$ and by Foreign Merges $\isforeign{R}
$,
so Totally Foreign Contents applies. $\qed$
so Totally Foreign Contents applies. $\qed$