complexity

aplas21/entail-hls.tex

@@ -54,18 +54,18 @@ The general case is dealt with in Section~\ref{ssec:ent-all};
 We show how to derive the coNP upper bound from the aforementioned decision procedure.  We first observe the following facts.
 \begin{enumerate}
 \item The original entailment problem is valid iff all of at most exponentially many 
-\emph{ordered entailment problems} are valid, since there are exponentially many total preorders.
-\item For each ordered entailment problem, it is valid iff all of at most exponentially many \emph{special ordered entailment problems}, where the consequent contains only one spatial atom, are valid.
+\emph{ordered entailment problems} are valid, since there are at most exponentially many total preorders.
+\item Each ordered entailment problem is valid iff all of at most exponentially many \emph{special ordered entailment problems}, where the consequent contains only one spatial atom, are valid.
 %\item The original entailment problem is invalid iff there is an invalid special ordered entailment problem instance.
 \item The special ordered entailment problem is in coNP.
 \end{enumerate}
 Therefore,  to show that $\varphi \models \psi$ is not valid, we 
 \begin{itemize}
-\item first guess a total preorder and obtain an ordered entailment problem, 
-\item then guess for the ordered entailment problem one special ordered entailment problem where the consequent contains only one spatial atom, 
-\item and finally guess and verify in polynomial time a polynomial-size witness for the invalidity of the special ordered entailment problem. 
+\item first guess a total preorder and obtain an ordered entailment problem (of polynomial size), 
+\item then guess for the ordered entailment problem one special ordered entailment problem (of polynomial size) where the consequent contains only one spatial atom, 
+\item finally guess and verify in polynomial time a polynomial-size witness for the invalidity of the special ordered entailment problem. 
 \end{itemize}
-The above procedure runs in nondeterministic polynomial time, thus showing the invalidity of the entailment problem is in NP. From this, we conclude that the entailment problem of {\slah} is in coNP.
+The above procedure runs in nondeterministic polynomial time, thus showing the invalidity of the entailment problem is in NP. From this, we deduce that the entailment problem of {\slah} is in coNP.
 
 %At first, it is easy to observe that if $\abs(\varphi)$ is unsatisfiable, 
 %then the entailment holds. Moreover, if $\abs(\varphi) \not\models \abs(\psi)$,