\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}
\[ \eqn{ Ingredients }{
\bigforall_{E \in \pendsof{X}{\py}} E \le Y
\right]
}\]
+\[ \eqn{ Suitable Tips }{
+ \bigforall_{\p \neq \patchof{L}, \; C \haspatch \p}
+ \bigexists_T
+ \pendsof{J}{\py} = \{ T \}
+ \land
+ \forall_{E \in \pendsof{K}{\py}} T \ge E
+ , \text{where} \{J,K\} = \{L,R\}
+}\]
\[ \eqn{ Foreign Merges }{
\patchof{L} = \bot \implies \patchof{R} = \bot
}\]
merge any Topbloke-controlled branch into any plain git branch.
Given those conditions, Tip Merge and Merge Acyclic do not apply.
-And by Foreign Contents for (wlog) Y, $\forall_{\p, D \in \py} D \not\le Y$
-so then by No Replay $D \not\isin Y$
-so $\neg [ Y \haspatch \p ]$ so neither
-Merge Ends condition applies.
+By Foreign Contents of $L$, $\patchof{M} = \bot$ 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
+conditions are satisifed.
So a plain git merge of non-Topbloke branches meets the conditions and
is therefore consistent with our model.
\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 $L
-\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch for $L$). So $D \neq C$.
+\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. $C \nothaspatch \p$.
+OK.
\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 $D \isin Y \equiv D \le Y$.)
-Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \zhaspatch \p$.
For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
\equiv D \isin L \lor D \isin R$.
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 $C \haspatch \p$.
+so $D \le C$. OK for $C \zhaspatch \p$.
Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
By $\merge$, $D \not\isin C$.
And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
-OK for $C \haspatch \p$.
+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 $D \le C$.
-OK for $C \haspatch \p$.
+OK for $C \zhaspatch \p$.
-So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+So, in all cases, $C \zhaspatch \p$.
+And by $L \haspatch \p$, $\exists_{F \in \py} F \le L$
+and this $F \le C$ so indeed $C \haspatch \p$.
\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 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.
-
-Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+various cases that
+if $M \haspatch \p$, $D \not\isin C$,
+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 $C \in \py, L \in \py$.
+By Tip Own Contents, $L \haspatch \p$ so $L \neq X$,
+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 $\merge$, $D \isin C$,
+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 $D \neq C, M \nothaspatch P, D \not\isin Y$:
+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 $D \neq C, M \haspatch P, D \isin Y$:
+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 $D \neq C, M \haspatch P, D \not\isin Y$:
+Consider $D \neq C, M \haspatch \p, D \not\isin Y$:
By $\merge$, $D \not\isin C$. OK.
$\qed$
\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 $\merge$, $D \isin
+$D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
+$D \isin L \equiv D \isin M$. So by $\merge$, $D \isin
C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
$D \isin L \equiv D \isin \baseof{L}$ and
$D \isin R \equiv D \isin \baseof{R}$.
+Apply Tip Merge condition.
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 $\merge$,
+So $D \isin M \equiv D \isin L$ so 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.
+$\baseof{C} = \baseof{R}$.
+Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
+
+$\qed$
+
+\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.
$\qed$
~ian / topbloke-formulae.git / blobdiff