formatting move/add some qeds

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 70d6d88200ffa76188a4e3fb553043e4094288a2..fda5026a8de1e76d083b9482b537e3d81915147a 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -369,7 +369,10 @@ Topbloke strips the metadata when exporting.
 Trivial.
 
 \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 $A$:
@@ -396,7 +399,9 @@ 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}
 
@@ -465,6 +470,7 @@ Used for removing a branch dependency.
 
 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.)
@@ -542,6 +548,8 @@ So $L \nothaspatch \p \implies C \nothaspatch \p$.
 Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
 so $L \haspatch \p \implies C \haspatch \p$.
 
+$\qed$
+
 \section{Foreign Inclusion}
 
 Consider some $D$ s.t. $\patchof{D} = \bot$.  $D \neq C$.
@@ -549,7 +557,9 @@ So by Desired Contents $D \isin C \equiv D \isin L$.
 By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
 
 And $D \le C \equiv D \le L$.
-Thus $D \isin C \equiv D \le C$.  $\qed$
+Thus $D \isin C \equiv D \le C$.
+
+$\qed$
 
 \section{Merge}
 
@@ -727,7 +737,9 @@ $C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
 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}