\section{Merge}
+\label{commit-merge}
Merge commits $L$ and $R$ using merge base $M$:
\gathbegin
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- \mergeof{C}{L}{M}{R}
+ \commitmergeof{C}{L}{M}{R}
\end{gather}
We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\text{otherwise} : & \false
\end{cases}
}\]
+\[ \eqn{ Base Merge }{
+ L \in \pn \implies
+ \big[
+ R \in \pn
+ \lor
+ R \in \foreign
+ \lor
+ ( R \in \pqy \land \pq \neq \p )
+ \big]
+}\]
\[ \eqn{ Merge Acyclic }{
L \in \pn
\implies
\right]
}\]
\[ \eqn{ Suitable Tips }{
- \bigforall_{\p \neq \patchof{L}, \; C \haspatch \p}
+ \bigforall_{\p \patchisin C, \; \py \neq \patchof{L}}
\bigexists_T
\pendsof{J}{\py} = \{ T \}
\land
\forall_{E \in \pendsof{K}{\py}} T \ge E
, \text{where} \{J,K\} = \{L,R\}
}\]
-\[ \eqn{ Foreign Merges }{
+\[ \eqn{ Foreign Merge }{
\isforeign{L} \implies \isforeign{R}
}\]
\subsection{Non-Topbloke merges}
We require both $\isforeign{L}$ and $\isforeign{R}$
-(Foreign Merges, above).
+(Foreign Merge, 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 $L$, $\isforeign{M}$ as well.
-So by Foreign Contents for any $A \in \{L,M,R\}$,
+By Foreign Ancestry of $L$, $\isforeign{M}$ as well.
+So by Foreign Ancestry 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
conditions are satisifed.
\subsection{No Replay}
-By definition of $\merge$,
+By definition of \commitmergename,
$D \isin C \implies D \isin L \lor D \isin R \lor D = C$.
So, by Ingredients,
Ingredients Prevent Replay applies. $\qed$
\subsection{Coherence and Patch Inclusion}
+$C$ satisfies
+\gathbegin
+ C \haspatch \p \lor C \nothaspatch \p
+\gathnext
+C \haspatch \p \equiv
+ \stmtmergeof{L \haspatch \p}{M \haspatch \p}{R \haspatch \p}
+\end{gather}
+which (given Coherence of $L$,$M$,$R$) is equivalent to
$$
\begin{cases}
L \nothaspatch \p \land R \nothaspatch \p : & C \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 Own Contents for $L$).
So $D \neq C$.
-Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+Applying \commitmergename\ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
OK.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
(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 \commitmergename, $D \isin C$. Also $D \le X$
so $D \le C$. OK for $C \zhaspatch \p$.
Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
-By $\merge$, $D \not\isin C$.
+By \commitmergename, $D \not\isin C$.
And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
OK for $C \zhaspatch \p$.
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 $D \le C$.
+Thus by \commitmergename, $D \isin C$. And $D \le Y$ so $D \le C$.
OK for $C \zhaspatch \p$.
So, in all cases, $C \zhaspatch \p$.
therefore we must have $L=Y$, $R=X$.
Conversely $R \not\in \py$
so by Tip Merge $M = \baseof{L}$. Thus $M \in \pn$ so
-by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
+by Base Acyclic $M \nothaspatch \p$. By \commitmergename, $D \isin C$,
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.
+$D \not\isin M$ so by \commitmergename, $D \isin C$. 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.
+Thus $D \not\le C$. By \commitmergename, $D \not\isin C$. 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.
+Thus $D \isin M$. By \commitmergename, $D \not\isin C$. OK.
Consider $D \neq C, M \haspatch \p, D \not\isin Y$:
-By $\merge$, $D \not\isin C$. OK.
+By \commitmergename, $D \not\isin C$. OK.
$\qed$
$D \neq C$. By Tip Contents of $L$,
$D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
-$D \isin L \equiv D \isin M$. So by $\merge$, $D \isin
+$D \isin L \equiv D \isin M$. So by \commitmergename, $D \isin
C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
-So $D \isin M \equiv D \isin L$ so by $\merge$,
+So $D \isin M \equiv D \isin L$ so by \commitmergename,
$D \isin C \equiv D \isin R$. But from Unique Base,
$\baseof{C} = \baseof{R}$.
Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\isforeign{D}$.
+Consider some $D \in \foreign$.
By Foreign Inclusion of $L, M, R$:
$D \isin L \equiv D \le L$;
$D \isin M \equiv D \le M$;
\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.
+R$. So by \commitmergename, $D \isin C$. And $D \le C$. OK.
\subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
-By $\merge$, $D \isin C$.
+By \commitmergename, $D \isin C$.
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 $\merge$, $D \not\isin C$.
+By \commitmergename, $D \not\isin C$.
And $D \not\le L, D \not\le R$ so $D \not\le C$.
OK
$\qed$
-\subsection{Foreign Contents}
+\subsection{Foreign Ancestry}
Only relevant if $\isforeign{L}$, in which case
-$\isforeign{C}$ and by Foreign Merges $\isforeign{R}$,
-so Totally Foreign Contents applies. $\qed$
+$\isforeign{C}$ and by Foreign Merge $\isforeign{R}$,
+so Totally Foreign Ancestry applies. $\qed$
+
+\subsection{Bases' Children}
+
+If $L \in \py, R \in \py$: not applicable for either $D=L$ or $D=R$.
+
+If $L \in \py, R \in \pn$: not applicable for $L$, OK for $R$.
+
+Other possibilities for $L \in \py$ are excluded by Tip Merge.
+
+If $L \in \pn, R \in \pn$: satisfied for both $L$ and $R$.
+
+If $L \in \pn, R \in \foreign$: satisfied for $L$, not applicable for
+$R$.
+
+If $L \in \pn, R \in \pqy$: satisfied for $L$, not applicable for
+$R$.
+
+Other possibilities for $L \in \pn$ are excluded by Base Merge.
+
+If $L \in \foreign$: not applicable for $L$; nor for $R$, by Foreign Merge.
+
+$\qed$