\newcommand{\py}{\pay{P}}
\newcommand{\pn}{\pan{P}}
+\newcommand{\pq}{\pa{Q}}
+\newcommand{\pqy}{\pay{Q}}
+\newcommand{\pqn}{\pan{Q}}
+
\newcommand{\pr}{\pa{R}}
\newcommand{\pry}{\pay{R}}
\newcommand{\prn}{\pan{R}}
\section{Notation}
-xxx make all conditions be in conditions, not in (or also in) intro
-
\begin{basedescript}{
\desclabelwidth{5em}
\desclabelstyle{\nextlinelabel}
\item[ $ C \ge D $ ]
$C$ is a descendant of $D$ in the git commit
-graph. This is a partial order, namely the transitive closure of
+graph. This is a partial order, namely the transitive closure of
$ D \in \set X $ where $ C \hasparents \set X $.
\item[ $ C \has D $ ]
None of these sets overlap. Hence:
\item[ $ \patchof{ C } $ ]
-Either $\p$ s.t. $ C \in \p $, or $\bot$.
+Either $\p$ s.t. $ C \in \p $, or $\bot$.
A function from commits to patches' sets $\p$.
\item[ $ \pancsof{C}{\set P} $ ]
-$ \{ A \; | \; A \le C \land A \in \set P \} $
+$ \{ A \; | \; A \le C \land A \in \set P \} $
i.e. all the ancestors of $C$
which are in $\set P$.
\item[ $ \pendsof{C}{\set P} $ ]
$ \{ E \; | \; E \in \pancsof{C}{\set P}
\land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}}
- E \neq A \land E \le A \} $
+ E \neq A \land E \le A \} $
i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$.
\item[ $ \baseof{C} $ ]
\item[ $ C \nothaspatch \p $ ]
$\displaystyle \bigforall_{D \in \py} D \not\isin C $.
-~ Informally, $C$ has none of the contents of $\p$.
+~ Informally, $C$ has none of the contents of $\p$.
Non-Topbloke commits are $\nothaspatch \p$ for all $\p$. This
includes commits on plain git branches made by applying a Topbloke
patch. If a Topbloke
patch is applied to a non-Topbloke branch and then bubbles back to
-the relevant Topbloke branches, we hope that
+the relevant Topbloke branches, we hope that
if the user still cares about the Topbloke patch,
git's merge algorithm will DTRT when trying to re-apply the changes.
(D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
\text{otherwise} : & D \not\isin M
\end{cases}
-$
+$
\end{basedescript}
\newpage
\[\eqn{Foreign Inclusion:}{
\bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
}\]
+\[\eqn{Foreign Contents:}{
+ \bigforall_{C \text{ s.t. } \patchof{C} = \bot}
+ D \le C \implies \patchof{D} = \bot
+}\]
\section{Some lemmas}
\text{as above with L and R exchanged}
\end{cases}
}\]
-\proof{
- Truth table xxx
-
- Original definition is symmetrical in $L$ and $R$.
+\proof{ ~ Truth table (ordered by original definition): \\
+ \begin{tabular}{cccc|c|cc}
+ $D = C$ &
+ $\isin L$ &
+ $\isin M$ &
+ $\isin R$ & $\isin C$ &
+ $L$ vs. $R$ & $L$ vs. $M$
+ \\\hline
+ y & ? & ? & ? & y & ? & ? \\
+ n & y & y & y & y & $\equiv$ & $\equiv$ \\
+ n & y & n & y & y & $\equiv$ & $\nequiv$ \\
+ n & n & y & n & n & $\equiv$ & $\nequiv$ \\
+ n & n & n & n & n & $\equiv$ & $\equiv$ \\
+ n & y & y & n & n & $\nequiv$ & $\equiv$ \\
+ n & n & y & y & n & $\nequiv$ & $\nequiv$ \\
+ n & y & n & n & y & $\nequiv$ & $\nequiv$ \\
+ n & n & n & y & y & $\nequiv$ & $\equiv$ \\
+ \end{tabular} \\
+ And original definition is symmetrical in $L$ and $R$.
}
\[ \eqn{Exclusive Tip Contents:}{
- \bigforall_{C \in \py}
+ \bigforall_{C \in \py}
\neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C )
\Bigr]
}\]
( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
\lor D = C
}\]
+\proof{ ~ Trivial.}
\[ \eqn{Transitive Ancestors:}{
\left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv
\proof{
The implication from right to left is trivial because
$ \pends() \subset \pancs() $.
-For the implication from left to right:
+For the implication from left to right:
by the definition of $\mathcal E$,
for every such $A$, either $A \in \pends()$ which implies
$A \le M$ by the LHS directly,
commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
by the LHS. And $A \le A''$.
}
+
\[ \eqn{Calculation Of Ends:}{
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
+ \begin{cases}
+ C \in \p : & \{ C \}
+ \\
+ C \not\in \p : & \displaystyle
\left\{ E \Big|
- \Bigl[ \Largeexists_{A \in \set A}
+ \Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
- \Bigl[ \Largenexists_{B \in \set A}
+ \Bigl[ \Largenexists_{B \in \set A}
E \neq B \land E \le B \Bigr]
\right\}
+ \end{cases}
}\]
-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.
+xxx proof tbd
+
+\[ \eqn{Ingredients Prevent Replay:}{
+ \left[
+ {C \hasparents \set A} \land
+ \\
+ \left(
+ D \isin C \implies
+ D = C \lor
+ \Largeexists_{A \in \set A} D \isin A
+ \right)
+ \right] \implies \left[
+ D \isin C \implies D \le C
+ \right]
+}\]
+\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 $D \le C$.
+}
-\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
- \land D \isin M$:}
-$D \not\isin C$. OK.
+\[ \eqn{Simple Foreign Inclusion:}{
+ \left[
+ C \hasparents \{ L \}
+ \land
+ \bigforall_{D} D \isin C \equiv D \isin L \lor D = C
+ \right]
+ \implies
+ \bigforall_{D \text{ s.t. } \patchof{D} = \bot}
+ D \isin C \equiv D \le C
+}\]
+\proof{
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+If $D = C$, trivially true. For $D \neq C$,
+by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
-$\qed$
+\[ \eqn{Totally Foreign Contents:}{
+ \bigforall_{C \hasparents \set A}
+ \left[
+ \patchof{C} = \bot \land
+ \bigforall_{A \in \set A} \patchof{A} = \bot
+ \right]
+ \implies
+ \left[
+ D \le C
+ \implies
+ \patchof{D} = \bot
+ \right]
+}\]
+\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}
\gathnext
\baseof{C}, \text{ if } C \in \py
\gathnext
- \bigforall_{\pa{Q}}
- \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
+ \bigforall_{\pq}
+ \text{ either } C \haspatch \pq \text{ or } C \nothaspatch \pq
\gathnext
- \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+ \bigforall_{\pqy \not\ni C} \pendsof{C}{\pqy}
\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.
+
+$C \haspatch \pq$ or $\nothaspatch \pq$ is represented as the
+set $\{ \pq | C \haspatch \pq \}$. Whether $C \haspatch \pq$
+is in stated
+(in terms of $I \haspatch \pq$ or $I \nothaspatch \pq$
+for the ingredients $I$),
+in the proof of Coherence for each kind of commit.
+
+$\pendsof{C}{\pq^+}$ 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 $A$:
Substitute into the contents of $C$:
\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A )
\lor D = C \]
-Since $D = C \implies D \in \py$,
+Since $D = C \implies D \in \py$,
and substituting in $\baseof{C}$, this gives:
\[ D \isin C \equiv D \isin \baseof{C} \lor
(D \in \py \land D \le A) \lor
\subsection{Base Acyclic}
-Need to consider only $A, C \in \pn$.
+Need to consider only $A, C \in \pn$.
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}
\subsubsection{For $A \nothaspatch P$:}
-Firstly, $C \not\in \py$ since if it were, $A \in \py$.
+Firstly, $C \not\in \py$ since if it were, $A \in \py$.
Thus $D \neq C$.
Now by contents of $A$, $D \notin A$, so $D \notin C$.
\[ A \nothaspatch P \implies C \nothaspatch P \]
$\qed$
-\subsection{Foreign inclusion:}
+\subsection{Foreign Inclusion:}
+
+Simple Foreign Inclusion applies. $\qed$
+
+\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} = \pqn
+\gathnext
+ D \isin B \equiv D \isin L \lor D = B
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{L} = \pa{L} \lor \patchof{L} = \bot
+}\]
+\[ \eqn{ Create Acyclic }{
+ L \not\haspatch \pq
+}\]
+
+\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 $D = B$, $D \in \pqn$.
+If $D \neq B$, $D \isin L$, and by Create Acyclic
+$D \not\in \pqy$. $\qed$
+
+\subsection{Coherence and Patch Inclusion}
-If $D = C$, trivial. For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
+Consider some $D \in \p$.
+$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$
+and $D \le B \equiv D \le L$.
+
+Thus $L \haspatch \p \implies B \haspatch P$
+and $L \nothaspatch \p \implies B \nothaspatch P$.
+
+$\qed$.
+
+\subsection{Foreign Inclusion}
+
+Simple Foreign Inclusion applies. $\qed$
+
+\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} = \pqy
+\gathnext
+ D \isin C \equiv D \isin B \lor D = C
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{B} = \pqn
+}\]
+\[ \eqn{ No Sneak }{
+ \pendsof{B}{\pqy} = \{ \}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. $\qed$
+
+\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 $B$, $D \isin B \implies D \not\in \pqy$.
+So $D \isin C \equiv D \isin \baseof{B}$.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+Not applicable.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+ \p = \pq \lor B \haspatch \p : & C \haspatch \p \\
+ \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+
+\proofstarts
+~ Consider some $D \in \py$.
+
+\subsubsection{For $\p = \pq$:}
+
+By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
+By No Sneak, $D \le B \equiv D = C$. Thus $C \haspatch \pq$.
+
+\subsubsection{For $\p \neq \pq$:}
+
+$D \neq C$. So $D \isin C \equiv D \isin B$,
+and $D \le C \equiv D \le B$.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Simple Foreign Inclusion applies. $\qed$
+
+\subsection{Foreign Contents}
+
+Not applicable.
\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}
By Currently Included, $D \isin L$.
By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by
-by Unique Tip, $D \le R^+ \equiv D \le L$.
+by Unique Tip, $D \le R^+ \equiv D \le L$.
So $D \isin R^+$.
By Base Acyclic, $D \not\isin R^-$.
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 $L$, $D \isin L \equiv D \le L$.
-So $D \isin C \equiv D \le L$.
-xxx up to here
+And $D \le C \equiv D \le L$.
+Thus $D \isin C \equiv D \le C$.
+
+$\qed$
-By Tip Contents of $R^+$, $D \isin R^+ \equiv D \isin \baseof{R^+}$
-i.e. $\equiv D \isin R^-$.
-So by $\merge$, $D \isin C \equiv D \isin L$.
+\subsection{Foreign Contents}
-Thus $D \isin C \equiv $
+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}
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,
But by Tip Merge condition on $\baseof{R}$,
$A \le \baseof{L} \implies A \le \baseof{R}$, so
$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.
That is, $\baseof{C} = \baseof{R}$.
\subsubsection{For $R \in \pn$:}
By Tip Merge condition on $R$ 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 $A \le C \equiv A \le R$.
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.
+Thus $A \le C \equiv A \le R$.
That is, $\baseof{C} = R$.
$\qed$
\subsection{Coherence and Patch Inclusion}
Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
-This involves considering $D \in \py$.
+This involves considering $D \in \py$.
\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
- \equiv D \isin L \lor D \isin R$.
+ \equiv D \isin L \lor D \isin R$.
(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 $\merge$, $D \isin C$. Also $D \le X$
so $D \le C$. OK for $C \haspatch \p$.
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 $D \not\le C$.
+By $\merge$, $D \not\isin C$.
+And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
OK for $C \haspatch \p$.
Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
-$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
+$D \not\le X$ 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 $C \haspatch \p$.
\proofstarts
-One of the Merge Ends conditions applies.
+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
$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK.
Consider $D \neq C, M \nothaspatch P, D \isin Y$:
-$D \le Y$ so $D \le C$.
+$D \le Y$ so $D \le C$.
$D \not\isin M$ so by $\merge$, $D \isin C$. OK.
Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
$D \not\le Y$. If $D \le X$ then
-$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
+$D \in \pancsof{X}{\py}$, 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 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}
-We need worry only about $C \in \py$.
+We need worry only about $C \in \py$.
And $\patchof{C} = \patchof{L}$
so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
of $C$, and its Coherence and Patch Inclusion, as just proved.
$\qed$
+\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