Accumulated small tweaks

diff --git a/strautintro.tex b/strautintro.tex
index f465637b50e168c5d45a693d5cde8ee87a72071d..eb33f2e9b329aed49e221cdbaa69eb7805fd9b12 100644 (file)
--- a/strautintro.tex
+++ b/strautintro.tex
@@ -49,7 +49,7 @@ transition process.}
 Now a run of our nondeterministic machine looks like $C_0 s_1 C_1 \ldots s_n
 C_n$ where each $C_i \subseteq \config$ and $c' \in C_{i+1}$ iff $\exists_{c
 \in C_i} . c' \in \delta\paren{c,s_{i+1}}$.  We consider the string
-$\vec{s}$ accepted by the automaton if $c \in C_n \cap \config_F$;
+$\vec{s}$ accepted by the automaton if $C_n \cap \config_F \ne \emptyset$;
 equivalently, we may say that a string is accepted if {\em there exists} a
 deterministic run, as defined earlier, where each $c_i \in C_i$, and $c_n
 \in \config_F$.
@@ -130,8 +130,8 @@ Here, we use the common table to describe itself.  Given a string $s \in
   compl={Find $B$ s.t. $\alang{B} = \alphabet^* \setminus \alang{A}$},
   relcompl={Find $B$ s.t. $\alang{B} = \alang{A} \setminus \alang{A'}$},
 %
-  intersect={Find $B$ s.t. $\alang{B} = \alang{A} \cap \alang{A'}$?},
-  union={Find $B$ s.t.  $\alang{B} = \alang{A} \cup \alang{A'}$?},
+  intersect={Find $B$ s.t. $\alang{B} = \alang{A} \cap \alang{A'}$},
+  union={Find $B$ s.t.  $\alang{B} = \alang{A} \cup \alang{A'}$},
 %
   hom={Find $B$ for hom. $h$ s.t. $\alang{B} = h\paren{\alang{A}}$},
   invhom={Find $B$ for hom. $h$ s.t. $\alang{A} = h\paren{\alang{B}}$},
@@ -153,11 +153,16 @@ Here, we use the common table to describe itself.  Given a string $s \in
 Of course, not every row in the table is independent of the others.  The
 following implications (at least) hold in all cases:
 \begin{itemize}
-       \item Subset testing implies equivalence testing.
-       \item Relative complement closure implies general complement closure.
-       \item Intersection and general complement closures implies
-          relative complement closure.  If, additionally, a class has
-          a decidable emptiness test, then it has a decidable subset test.
+       \item Subset testing implies equivalence testing, of which universality
+          and emptiness testing are often special cases (some automata
+          families may not be able to represent $\emptyset$ or
+          $\alphabet^*$).
+       \item Relative complement closure implies general complement closure, assuming $\alphabet^*$ is expessible.
+       \item Intersection and general complement closures imply
+          relative complement closure.
+    \item If a class has relative complement closure and a decidable emptiness test,
+          then it has a decidable subset test:
+          $S \subseteq T \Leftrightarrow S \setminus T = \emptyset$.
        \item Closure under arbitrary homomorphism implies closure under
           $\epsilon$-free homomorphisms.
 \end{itemize}

diff --git a/zoo-tree/ra.tex b/zoo-tree/ra.tex
index bef0a55d36acf20db78bbd3d5344a576313ffbf1..942cdd995f3bce8b88008c513fc88f321e19ee55 100644 (file)
--- a/zoo-tree/ra.tex
+++ b/zoo-tree/ra.tex
@@ -22,11 +22,10 @@ C]{jacquemard:tamodeq} and in contradiction to the statement made in
 {\em deterministic} and {\em complete} RAs, carries over to the
 non-deterministic case as well.}
 %
-that emptiness is not decidable for the class of nondeterminstic RAs.  (This
-proof generalizes to a large class of tree automata; see
-\autoref{sec:tree-sepex:twocm}.)  We feel that a few words said about this
+that emptiness is not decidable for the class of nondeterminstic RAs.  We feel that a few words said about this
 proof here may help to de-mystify it.  A decomposition if this machine into
-simpler automata whose intersection is this machine may be found in
+simpler automata whose intersection is this machine, which demonstrates its
+applicability to a large collection of automata classes, may be found in
 \ref{sec:tree-sepex:twocm}.
 
 \begin{wrapfigure}{r}{3.5in}\centering\begin{tikzpicture}
@@ -63,9 +62,9 @@ of the machine's execution, denoted $C_i$, and h$_{i-1}$.  The equality
 constraints are introduced only at the top and at ``g'' nodes in the tree;
 the state labels are carefully constructed so that only three ranks are
 required to satisfy the Reduction metaconstraint: a base rank, which
-includes all labels in the $C_i$ trees is $\le$ a ``g''-state rank which
+includes all labels in the $C_i$ trees, is $\le$ a ``g''-state rank, which
 includes the labels of all ``g'' and ``h'' nodes in the tree except the
-apex; the accepting state at the root is the sole member of the third rank.
-There will be $2^{n-i}$ copies of $h_i$ (or $C_i$) in a tree encoding $n$
-steps.  This construction is {\em deeply} dependent on the fact that
+apex, and the accepting state at the root is the sole member of the third
+rank.  There will be $2^{n-i}$ copies of $h_i$ (or $C_i$) in a tree encoding
+$n$ steps.  This construction is {\em deeply} dependent on the fact that
 non-determinism allows equal trees to have several distinct runs.