X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=fd45642df697518317032ed9c9956c62baf7c1dd;hp=e9b5aa26c7272d039d3f14a33dfb098563c466dd;hb=57f83cd8bc6bcf23e45739c00c83c9a8672ae701;hpb=65959570cc03ee3de82eefcb2dc51d7ae9f3aebc

diff --git a/merge.tex b/merge.tex
index e9b5aa2..fd45642 100644
--- a/merge.tex
+++ b/merge.tex
@@ -47,7 +47,7 @@ $L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
    \right]
 }\]
 \[ \eqn{ Foreign Merges }{
-    \patchof{L} = \bot \equiv \patchof{R} = \bot
+    \patchof{L} = \bot \implies \patchof{R} = \bot
 }\]
 
 \subsection{Non-Topbloke merges}
@@ -59,7 +59,9 @@ 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
+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.
 
 So a plain git merge of non-Topbloke branches meets the conditions and
@@ -79,7 +81,7 @@ and calculate $\pendsof{C}{\pn}$.  So we will consider some
 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,
@@ -98,7 +100,7 @@ That is, $\baseof{C} = \baseof{R}$.
 
 \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 $A \le C \equiv A \le R$.
@@ -156,27 +158,29 @@ 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 $D = C$:  Thus $C \in \py, L \in \py$.  By Tip Contents
+for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
+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.
 
 $\qed$
@@ -216,8 +220,8 @@ $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
 \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.
 
@@ -229,14 +233,16 @@ By Tip Contents
 $D \isin L \equiv D \isin \baseof{L}$ 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.
 
 $\qed$