foreign notation: change \bot to \foreign everywhere

[topbloke-formulae.git] / anticommit.tex

diff --git a/anticommit.tex b/anticommit.tex
index cf93fd2998cc593323055552dc92db7c4c11d8e4..72469b80d69fa67aa410381faf5768262d15964e 100644 (file)
--- a/anticommit.tex
+++ b/anticommit.tex
@@ -19,7 +19,7 @@ R^+ \in \pry \land R^- = \baseof{R^+}
 \[ \eqn{ Into Base }{
  L \in \pln
 }\]
 \[ \eqn{ Into Base }{
  L \in \pln
 }\]

-\[ \eqn{ Unique Tip }{
+\[ \eqn{ Correct Tip }{

  \pendsof{L}{\pry} = \{ R^+ \}
 }\]
 \[ \eqn{ Currently Included }{
  \pendsof{L}{\pry} = \{ R^+ \}
 }\]
 \[ \eqn{ Currently Included }{


@@ -28,7 +28,7 @@ R^+ \in \pry \land R^- = \baseof{R^+}
 
 \subsection{Ordering of Ingredients:}
 
 
 \subsection{Ordering of Ingredients:}
 

-By Unique Tip, $R^+ \le L$.  By definition of $\base$, $R^- \le R^+$
+By Correct Tip, $R^+ \le L$.  By definition of $\base$, $R^- \le R^+$

 so $R^- \le L$.  So $R^+ \le C$ and $R^- \le C$.
 $\qed$
 
 so $R^- \le L$.  So $R^+ \le C$ and $R^- \le C$.
 $\qed$
 


@@ -62,7 +62,7 @@ $D \not\isin R^-$.  Thus $D \not\isin C$.  OK.
 By Currently Included, $D \isin L$.
 
 By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but
 By Currently Included, $D \isin L$.
 
 By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but

-by Unique Tip, $D \le R^+ \equiv D \le L$.
+by Correct Tip, $D \le R^+ \equiv D \le L$.

 So $D \isin R^+$.
 
 By Base Acyclic for $R^-$, $D \not\isin R^-$.
 So $D \isin R^+$.
 
 By Base Acyclic for $R^-$, $D \not\isin R^-$.


@@ -125,9 +125,13 @@ OK.
 
 $\qed$
 
 
 $\qed$
 

+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies.  $\qed$
+

 \subsection{Foreign Inclusion}
 
 \subsection{Foreign Inclusion}
 

-Consider some $D$ s.t. $\patchof{D} = \bot$.  $D \neq C$.
+Consider some $D$ s.t. $\patchof{D} = \foreign$.  $D \neq C$.

 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 by Desired Contents $D \isin C \equiv D \isin L$.
 By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.