@@ -189,9 +189,9 @@ We build a heap $h$ from $s$ and $\varphi$ as follows:

%

\item For every predicate atom $\hls{}(t_1,t_2; t_3)$ in $\varphi$, we have $s \models\abs(\hls{}(t_1,t_2; t_3))$. Then either $s(t_1)= s(t_2)$ or $s(t_1) < s(t_2)$ and $s \models\abs^+(\hls{}(x, y; z))[t_1/x, t_2/y, t_3/z]$.

\begin{itemize}

\item For the formal case, let $h(s(t_1))$ undefined.

\item For the first case, we let $h(s(t_1))$ undefined.

\item For the latter case, $s \models(2= t_3\wedge t_1 < t_2\wedge t_2- t_1\equiv0\bmod2)\vee(2 < t_3\wedge2\le t_2- t_1)$.

\item For the second case, $s \models(2= t_3\wedge t_1 < t_2\wedge t_2- t_1\equiv0\bmod2)\vee(2 < t_3\wedge2\le t_2- t_1)$.

\begin{itemize}

\item If $s(t_3)=2$, then let $h(s(t_1)+2(i-1))=2$ and $h(s(t_1)+2i-1)=1$ for every $i: 1\le i \le\frac{s(t_2)- s(t_1)}{2}$.

\item If $s(t_3) > 2$, then from $s(t_2)- s(t_1)\ge2$, we know that there is a sequence of numbers $n_1, \cdots, n_\ell$ such that $2\le n_i \le s(t_3)$ for every $i \in[\ell]$ and $s(t_2)= s(t_1)+\sum\limits_{i \in[\ell]} n_i$. We then let $h(s(t_1)+\sum\limits_{j \in[i-1]} n_j)= n_i$ for every $i \in[\ell]$, and let $h(n')=1$ for all the other addresses $n' \in[s(t_1), s(t_2)-1]$.

@@ -49,7 +49,12 @@ In Section~\ref{ssec:ent-1}, we consider the special case

The general case is dealt with in Section~\ref{ssec:ent-all};

the procedure calls the special case for the first atom of the consequent with

all the compatible prefixes of the antecedent, and

it does a recursive call for the remainders of the consequent and the antecedent. Note that to find all the compatible prefixes of the antecedent, some spatial atoms in the antecedent might be split into several ones. The coNP upper bound is explained in the sequel: 1) the original entailment problem is reduced to at most exponentially many ordered entailment problems since there are exponentially total preorders, 2) each ordered entailment problem can be reduced further to exponentially many special ordered entailment problems where there is one spatial atom in the consequent, 3) the original entailment problem is invalid iff there is an invalid special ordered entailment problem instance, 4) the special ordered entailment problem is in coNP.

it does a recursive call for the remainders of the consequent and the antecedent. Note that to find all the compatible prefixes of the antecedent, some spatial atoms in the antecedent might be split into several ones.

The coNP upper bound is explained in the sequel:

1) the original entailment problem is reduced to at most exponentially many ordered entailment problems since there are exponentially many total preorders,

2) each ordered entailment problem can be reduced further to exponentially many special ordered entailment problems where there is one spatial atom in the consequent,

3) the original entailment problem is invalid iff there is an invalid special ordered entailment problem instance,

4) the special ordered entailment problem is in coNP.

%At first, it is easy to observe that if $\abs(\varphi)$ is unsatisfiable,

%then the entailment holds. Moreover, if $\abs(\varphi) \not\models \abs(\psi)$,

@@ -27,8 +27,9 @@ we prefer to keep it in the signature of \PbA\ to obtain quantifier free formula

\PbA\ is a useful tool for showing complexity classes because its

satisfiability problem belongs to various complexity classes depending

on the number of quantifier alternations~\cite{Haase2018ASG}.

In this paper, we consider quantifier-free \PbA\ formulas (abbreviated as \qfpa) and the $\mathsf{\Sigma}_1$-fragment of \PbA (abbreviated as \EPbA), which

contains existentially quantified Presburger arithmetic formulas. The satisfiability problem of {\qfpa} and {\EPbA} is NP-complete.

In this paper, we consider quantifier-free \PbA\ formulas (abbreviated as \qfpa) and the $\mathsf{\Sigma}_1$-fragment of \PbA\ (abbreviated as \EPbA), which