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 by No Replay for $A$, $D \not\isin A$.
+Thus $\pendsof{A}{\py} = \{ \}$ 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.
putative ancestor $A \in \pn$ and see whether $A \le C$.
By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
-But $C \in py$ and $A \in \pn$ so $A \neq C$.
+But $C \in \py$ and $A \in \pn$ so $A \neq C$.
Thus $A \le C \equiv A \le L \lor A \le R$.
By Unique Base of L and Transitive Ancestors,
\subsubsection{For $R \in \pn$:}
-By Tip Merge condition on $R$ and since $M \le R$,
+By Tip Merge condition and since $M \le R$,
$A \le \baseof{L} \implies A \le R$, so
$A \le R \lor A \le \baseof{L} \equiv A \le R$.
Thus $A \le C \equiv A \le R$.
\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 $\neg[ L \nothaspatch \p ]$ by Tip Self Inpatch for $L$).
+So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
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 $D = C$: Thus $C \in \py, L \in \py$.
+By Tip Self Inpatch, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
+therefore we must have $L=Y$, $R=X$.
+By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
+by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
+and $D \le C$, consistent with $C \haspatch \p$. OK.
-Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+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.
$\qed$
~ian / topbloke-formulae.git / blobdiff