text:
\section{Review of\ the Splitting Set Theorem and Hybrid ASP}

We will begin with a brief review of ASP. Let $At$ be a nonempty set of
symbols called \emph{atoms}.  A \emph{block} is an expression of the form
\begin{equation}
b_{1},...,b_{k},\;not\;b_{k+1},\;...,\;not\;b_{k+m}\label{block}%
\end{equation}
where $b_{1},\;...,\;b_{k+m}$ are atoms. For a block $B$ as above, let the
\emph{set of atoms of }$B$ be defined as $At\left(  B\right)  \equiv
\{b_{1},\;...,\;b_{k+m}\}$. $B^{+}\equiv b_{1},\;...,\;b_{k}$ is called the
\emph{positive part of }$B$, and $B^{-}\equiv$ $not\;b_{k+1}%
,\;...,\;not\;b_{k+m}$ is called the \emph{negative part of }$B$. A set
operation applied to a block $B$ will indicate the same set operation applied
to $At\left(  B\right)  $ with the block being reconstructed from the result
of the set operation. For instance $b_{1},\;b_{2},\;not\;b_{3},\;b_{4}%
\;\backslash\;\{b_{1},\;b_{4}\}$ will indicate a block $b_{2},\;not\;b_{3}$.

A \emph{normal propositional logic programming }rule is an expression of the
form%
\begin{equation}
r\equiv a:-\;B\label{clause2}%
\end{equation}
where $a$ is an atom and $B$ is a block. We define the \emph{head of }$r$ as
$head\left(  r\right)  \equiv a$, and we define the \emph{body of }$r$ as
$body\left(  r\right)  \equiv B$. We define $At\left(  r\right)
\equiv\left\{  a\right\}  \cup At\left(  B\right)  $.

\bigskip Given any set $M\subseteq At$ and a block $B$, we say that $M$
satisfies $B$, written $M\models B$, if $At\left(  B^{+}\right)  \subseteq M$
and $At\left(  B^{-}\right)  \cap M=\emptyset$. For a rule $r$, we say that
$M$ satisfies $r$, written $M\models r$, if whenever $M$ satisfies the body of
$r$, then $M$ satisfies the head of $r$. A \emph{normal logic program} $P$ is
a set of rules. We say that $M\subseteq At$ is a model of $P$, written
$M\models P$, if $M$ satisfies every rule of $P$.

A \emph{Horn rule} is the rule with the empty negative part. A Horn program
$P$ is a set of Horn rules. Each Horn program $P$ has a least model under
inclusion, $LM_{P}$, which can be defined using the \emph{one-step provability
operator} $T\left[  P\right]  $ as follows. For any set $A$, let
$\mathcal{P}\left(  A\right)  $ denote the set of all subsets of $A$. The
one-step provability operator $T\left[  P\right]  :\mathcal{P}\left(
At\right)  \rightarrow\mathcal{P}\left(  At\right)  $ associated with the Horn
program $P$ \cite{EmdenK76} is defined by setting
\[
T\left[  P\right]  (M)=M\cup\{a:\exists r\in P\;(a=head(r)\wedge M\models
body(r))\}
\]
for any $M\in\mathcal{P}\left(  At\right)  $. We define $T\left[  P\right]
^{n}(M)$ by induction by setting $T\left[  P\right]  ^{0}\left(  M\right)
=M$, $T\left[  P\right]  ^{1}(M)=T\left[  P\right]  (M)$ and $T\left[
P\right]  ^{n+1}(M)=T\left[  P\right]  (T\left[  P\right]  ^{n}(M))$. Then the
least model $LM_{P}$ can be computed as \linebreak $LM_{P}=\bigcup_{n\geq0}T\left[
P\right]  ^{n}(\emptyset).$

If $P$ is a normal logic program and $M\subseteq At$, then the
Gelfond-Lifschitz (GL)\ reduct of $P$ with respect to $M$ \cite{GelfondL88} is
the Horn program $P^{M}$ which results by eliminating those rules $r$ such
that $M\not \models body\left(  r\right)  ^{-}$ and replacing other rules $r$
by $head\left(  r\right)  :-body\left(  r\right)  ^{+}$. We then say that $M$
is a \emph{stable model} for $P$ if $M$ equals the least model of $P^{M}$.

An \emph{answer set programming rule }is an expression of the form
(\ref{clause2}) where $a,b_{1},\ldots,b_{k+m}$ are classical literals, i.e.,
either positive atoms or atoms preceded by the classical negation sign $\lnot
$. The set of literals of $At$ will be denoted $Lit_{At}$. \emph{Answer sets}
are defined in analogy to stable models, but taking into account that atoms
may be preceded by classical negation and that atoms $a$ and classically
negated atoms $\lnot a$ are mutually exclusive in answer sets.

We will now follow \cite{DBLP:conf/iclp/LifschitzT94} in review of the
Splitting Set Theorem and the Splitting Sequence Theorem. A \emph{splitting
set} for a program $P$ is any set $U\subseteq At$ such that for every rule
$r\in P$ if $head\left(  r\right)  \in U$ then $At\left(  r\right)  \subseteq
U$. The set of rules $r\in P$ such that $At\left(  r\right)  \subseteq U$ is
called the \emph{bottom }of $P$ relative to the splitting set $U$ and is
denoted by $b_{U}\left(  P\right)  $. The set $P\backslash b_{U}\left(
P\right)  $ is the \emph{top} of $P$ relative to $U$.

Consider $X\subseteq At$. For each rule $r\in P$ such that $At(body\left(
r\right)  ^{+})\cap U\subseteq X$ and $At(body\left(  r\right)  ^{-})\cap
U\cap X=\emptyset$ take the rule $r^{\prime}$ defined by%
\[
head\left(  r\right)  \;:-\;body\left(  r\right)  \;\backslash\;U
\]

The program consisting of all rules $r^{\prime}$ obtained in this way will be
denoted by $\epsilon_{U}\left(  P,X\right)  $.

A \emph{solution }to $P$ with respect to $U$ is a pair $\left(  X,Y\right)  $
of sets of literals such that

\begin{itemize}
\item $X$ is an answer set for $b_{U}\left(  P\right)  $

\item $Y$ is an answer set for $\epsilon_{U}\left(  P\;\backslash
\;b_{U}\left(  P\right)  ,\;X\right)  $

\item $X\cup Y$ is consistent (a set is consistent if for any atom $a$ it does
not contain both $a$ and classically negated atom $-a$)
\end{itemize}

\textbf{Splitting Set Theorem. }\emph{Let }$U$\emph{ be a splitting set for a
program }$P$\emph{. A set }$A$\emph{ of literals is a consistent answer set
for }$P$\emph{ if and only if }$A=X\cup Y$\emph{ for some solution }$\left(
X,Y\right)  $\emph{ to }$P$\emph{ with respect to }$U$\emph{.}

\bigskip

We will now review extending the definition of a splitting set to a splitting
sequence. A \emph{sequence} is a family whose index set is an initial segment
of ordinals, $\{\alpha:\alpha<\mu\}$. The ordinal $\mu$ is the \emph{length
}of the sequence. A sequence $\left\langle U_{\alpha}\right\rangle
_{\alpha<\mu}$ of sets is \emph{monotone} if $U_{\alpha}\subset U_{\beta}$
whenever $\alpha<\beta$, and \emph{continuous }if, for each limit ordinal
$\alpha<\mu$, $U_{\alpha}=%
{\displaystyle\bigcup\limits_{\beta<\alpha}}
U_{\beta}$.

A \emph{splitting sequence }for a program $P$ is a monotone, continuous
sequence $\left\langle U_{\alpha}\right\rangle _{\alpha<\mu}$ of splitting
sets for $P$ such that $%
{\displaystyle\bigcup\limits_{\alpha<\mu}}
U_{\alpha}=Lit_{At}$. The definition of a solution with respect to a splitting
set is extended to splitting sequence as follows. A \emph{solution} to $P$
with respect to $\left\langle U_{\alpha}\right\rangle _{\alpha<\mu}$ is a
sequence $\left\langle X_{\alpha}\right\rangle _{\alpha<\mu}$ of sets of
literals such that

\begin{itemize}
\item $X_{0}$ is an answer set for $b_{U_{0}}\left(  P\right)  $,

\item for any $\alpha$ such that $\alpha+1<\mu$, $X_{\alpha+1}$ is an answer
set for $\epsilon_{U_{\alpha}}(b_{U_{\alpha+1}}\left(  P\right)  \backslash
b_{U_{\alpha}}\left(  P\right)  ,\;%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta})$,

\item for any limit ordinal $\alpha<\mu$, $X_{\alpha}=\emptyset$,

\item $%
{\displaystyle\bigcup\limits_{\alpha<\mu}}
X_{\alpha}$ is consistent.
\end{itemize}

\textbf{Splitting Sequence Theorem.}\emph{ Let }$U\equiv\left\langle
U_{\alpha}\right\rangle _{\alpha<\mu}$\emph{ be a splitting sequence for a
program }$P$\emph{. A set }$A$\emph{ of literals is a consistent answer set
for }$P$\emph{ if and only if }$A=%
{\displaystyle\bigcup\limits_{\alpha<\mu}}
X_{\alpha}$\emph{ for some solution }$\left\langle X_{\alpha}\right\rangle
_{\alpha<\mu}$\emph{ to }$P$\emph{ with respect to }$U$\emph{.}

\bigskip

We will now proceed with the review of Hybrid ASP. A Hybrid ASP\ program $P$
has an underlying parameter space $S$. Elements of $S$ are of the form
$\mathbf{p}=(t,x_{1},\ldots,x_{l})$ where $t$ is time and $x_{i}$ are
arbitrary parameter values. We shall let $t(\mathbf{p})$ denote $t$ and
$x_{i}(\mathbf{p})$ denote $x_{i}$ for $i=1,\ldots,l$. We refer to the
elements of $S$ as \emph{generalized positions}. Let $At$ be a set of atoms of
$P$. Then the universe of $P$ is $At\times S$. Let $B$ be a block. We will
define%
\[
B\times\mathbf{p}\equiv\{\left(  x,\mathbf{p}\right)  :\;x\in B\}.
\]

If $M\subseteq At\times S$, we let $GP(M)=\{\mathbf{p}\in S:(\exists a\in
At)((a,\mathbf{p})\in M)\}$. Given an \emph{initial condition}, defined as a
subset  $I\subseteq S$ let $GP_{I}\left(  M\right)  =GP\left(  M\right)  \cup
I$. Given $M\subseteq At\times S$ and $\mathbf{p}\in S$, we say that $M$ and
initial condition $I$ satisfy a block $B$ of the form (\ref{block}) at the
generalized position $\mathbf{p}$, written $M\models_{I}\left(  B,\mathbf{p}%
\right)  $, if the following holds:

\begin{itemize}
\item if $B^{+}\neq\emptyset$ then $B^{+}\times\mathbf{p}\subseteq M$ and
$B^{-}\times\mathbf{p\cap M=\emptyset}$

\item if $B^{+}=\emptyset$ then $B^{-}\times\mathbf{p\cap M=\emptyset}$ and
$\mathbf{p}\in GP_{I}\left(  M\right)  $.
\end{itemize}

We say that $M$ satisfies a n-tuple of blocks written as $B_{1};\;...;\;B_{n}$
with the initial condition $I$ at the n-tuple of generalized positions
$(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})$, written $M\models_{I}%
(B_{1};\;...;B_{n},\;(\mathbf{p}_{1},\;...,\mathbf{p}_{n}))$, if $M\models
_{I}(B_{i},\mathbf{p}_{i})$ for $i=1,...,n$.

There are two types of rules in Hybrid ASP. \textbf{Advancing rules} are of
the form
\begin{equation}
r\equiv a:-B_{1};B_{2};\ldots;B_{n}:A,O\label{advrule}%
\end{equation}
where $A$ is a function returning a set of generalized positions, $body\left(  r\right)  \equiv B_{1},\;...,\;B_{n}$
are blocks, $head\left(  r\right)  \equiv a$ is a literal, and $O$ is a subset
of $S^{n}$ such that if $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})\in O$, then
$t(\mathbf{p}_{1})<\cdots<t(\mathbf{p}_{n})$ and $A\left(  \mathbf{p}%
_{1},\ldots,\mathbf{p}_{n}\right)  $ ($A$ applied to $ \mathbf{p}_{1},\ldots,\mathbf{p}_{n}$) is a subset of $S$ such that for all
$\mathbf{q}\in A\left(  \mathbf{p}_{1},\ldots,\mathbf{p}_{n}\right)  $,
$t(\mathbf{q})>t(\mathbf{p}_{n})$. Here and in the next rule, we allow blocks
to be empty for any $i$. $O$ is called the \emph{constraint set} of the rule
$r$ and will be denoted by $CS(r)$. $A$ is called the \emph{advancing
algorithm} of the rule $r$ and is denoted by $\operatorname*{Adv}(r)$. The
arity of rule $r$, $N\left(  r\right)  $, is equal to $n$.

The idea is that if $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})\in O$ and for each
$i$, $B_{i}$ is satisfied at the generalized position $\mathbf{p}_{i}$, then
the function $A$ can be applied to $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})$
to produce a set of generalized positions $O^{\prime}$ such that if
$\mathbf{q}\in O^{\prime}$, then $t(\mathbf{q})>t(\mathbf{p}_{n})$ and
$(a,\mathbf{q})$ holds. Thus advancing rules are like input-output devices in
that the function $A$ allows the user to derive possible successor
generalized positions as well as certain atoms $a$ which are to hold at such
positions. The advancing algorithm $A$ can access outside sources quite
arbitrarily in that it may involve functions for solving differential or
integral equations, solving a set of linear equations or linear programming
equations, solving an optimization problem, etc. (as for example in
\cite{HybridASP2}).

\textbf{Stationary rules} are of the form
\begin{equation}
r\equiv a:-B_{1};B_{2};\ldots;B_{n}:H,O\label{strule}%
\end{equation}
\newline where $body\left(  r\right)  \equiv B_{1},...,B_{n}$ are blocks,
$head\left(  r\right)  \equiv a$ is a literal, $H$ is called a \emph{boolean algorithm} of
the rule $r$ and will be denoted by $Bool\left(  r\right)  $, and $O\subseteq
S^{k}$ is the constraint set of the rule $r$ denoted $CS(r)$. A boolean algorithm is a function returning either true or false. We will
sometimes treat a boolean algorithm of the rule as a set. For instance $H\cap
O$ will indicate all the n-tuples of generalized positions $(\mathbf{p}%
_{1},\ldots,\mathbf{p}_{n})$ such that $H\left(  \mathbf{p}_{1},\ldots
,\mathbf{p}_{n}\right)  $ is true and $(\mathbf{p}_{1},\ldots,\mathbf{p}%
_{n})\in O$. The arity of rule $r$, $N\left(  r\right)  $, is equal to $n$.

Stationary rules are much like normal logic programming rules in that they
allow us to derive new atoms at a given generalized position $\mathbf{p}_{n}$.
The idea is that if $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})\in O\cap H$ and
for each $i$, $B_{i}$ is satisfied at the generalized position $\mathbf{p}%
_{i}$, then $(a,\mathbf{p}_{n})$ holds. The difference is that a derivation
with our stationary rules can depend on what happens in the multiple past time
points and the boolean algorithm $H$ can be any sort of a function which
returns either true or false. \ \newline

For an advancing rule or a stationary rule $r$ as above we define the\emph{
positive part of the body} of $r$, denoted $body\left(  r\right)  ^{+}\equiv
B_{1}^{+};...;B_{n}^{+}$ and we define the \emph{negative part of the body }of
$r$, denoted $body\left(  r\right)  ^{-}\equiv B_{1}^{-};...;B_{n}^{-}$. For
the rest of the paper, we denote by $n$ the arity of a hybrid ASP\ rule when
the rule is clear from the context.

A Hybrid ASP program $P$ is a collection of Hybrid ASP advancing and
stationary rules. To define the notion of a stable model of $P$, we first must
define the notion of a Hybrid ASP Horn program and the one-step provability
operator for Hybrid ASP Horn programs.

A \emph{Hybrid ASP Horn program} is a Hybrid ASP program which does not
contain any negated atoms. Let $P$ be a Horn Hybrid ASP\ program and
$I\subseteq S$ be an initial condition. Then the one-step provability operator
$T\left[  P,I\right]  $ is defined so that given $M\subseteq At\times S$,
$T\left[  P,I\right]  (M)$ consists of $M$ together with the set of all
$(a,J)\in At\times S$ such that

\begin{enumerate}
\item \noindent there exists a stationary rule $r$ and $(\mathbf{p}_{1}%
,\ldots,\mathbf{p}_{n})\in CS\left(  r\right)  \cap Bool\left(  r\right)
\cap\left(  GP_{I}(M)\right)  ^{n}$ such that \linebreak$(head\left(  r\right)
,J)=(a,\mathbf{p}_{n})$ and $M\models(body\left(  r\right)  ,\;(\mathbf{p}%
_{1},...,\mathbf{p}_{n}))$ or

\item there exists an advancing rule $r$ and $(\mathbf{p}_{1},\ldots
,\mathbf{p}_{n})\in CS\left(  r\right)  \cap\left(  GP_{I}(M)\right)  ^{n}$
such that\linebreak\ $J\in\operatorname*{Adv}\left(  r\right)  (\mathbf{p}%
_{1},\ldots,\mathbf{p}_{n})$ and $M\models(body\left(  r\right)
,\;(\mathbf{p}_{1},...,\mathbf{p}_{n}))$ and $a=head\left(  r\right)  $.
\end{enumerate}

The stable model semantics for Hybrid ASP programs is defined as follows. Let
$M\subseteq At\times S$ and $I$ be an initial condition in $S$. An Hybrid ASP
rule $r\equiv a:-B_{1};\ldots,B_{n}:A,O$ is \emph{inapplicable} for $(M,I)$ if
for all $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})\in O\cap\left(  GP_{I}%
(M)\right)  ^{n}$, either (i) there is an $i$ such that $M\not \models
(B_{i}^{-},\mathbf{p}_{i})$, (ii) $A\left(  \mathbf{p}_{1},\ldots
,\mathbf{p}_{n}\right)  \cap GP_{I}(M)=\emptyset\,$ if $A$ is an advancing
algorithm, or (iii) $A(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})=0$ if $A$ is a
boolean algorithm.

\bigskip If $r$ is not inapplicable for $\left(  M,I\right)  $ then we define
the GL\ reduct of $r$ over $M$ and $I$, denoted by $r^{M,I}$ as follows:

\begin{enumerate}
\item If $r$ is an advancing rule $r\equiv a:-B_{1};...;B_{n}:A,O$ then
$r^{M,I}\equiv B_{1}^{+};\ldots,B_{n}^{+}:A^{M,I},O^{M,I}$ where $O^{M,I}$ is
equal to the set of $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})$ in $O\cap\left(
GP_{I}(M)\right)  ^{n}$ such that \ $M\models_{I}(body\left(
r\right)  ^{-},\;(\mathbf{p}_{1},\ldots,\mathbf{p}_{n}))$ and $A(\mathbf{p}%
_{1},\ldots,\mathbf{p}_{n})\cap GP_{I}(M)\neq\emptyset$, and $A^{M,I}%
(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})\equiv A(\mathbf{p}_{1},\ldots
,\mathbf{p}_{n})\cap GP_{I}(M)$.

\item If $r$ is a stationary rule $r\equiv a:-B_{1};...;B_{n}:A,O$ then
$r^{M,I}\equiv a:-B_{1}^{+};\ldots,B_{n}^{+}:H|_{O^{M,I}},O^{M,I}$ where
$O^{M,I}$ is equal to the set of all $(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})$
in $O\cap\left(  GP_{I}(M)\right)  ^{n}$ such that \linebreak $\ M\models
_{I}(body\left(  r\right)  ^{-},\;(\mathbf{p}_{1},\ldots,\mathbf{p}_{n}))$ and
$H(\mathbf{p}_{1},\ldots,\mathbf{p}_{n})$ is true.
\end{enumerate}

\bigskip One note to make about the definition above is that GL\ reduct cannot
derive generalized positions that are not in $GP_{I}\left(  M\right)  $. This
is because the range of $A^{M,I}$ in the definition is restricted to
$GP_{I}(M)$.

We form a GL reduct of $P$ over $M$ and $I$, $P^{M,I}$ as follows.

\begin{enumerate}
\item \noindent Eliminate all rules which are inapplicable for $(M,I)$.

\item If a rule $r\in P$ is not eliminated in step 1, then replace it by the
rule $r^{M,I}$.
\end{enumerate}

We then say that $M$ is a\emph{ stable model of }$P$\emph{\ with initial
condition }$I$ if $%
{\displaystyle\bigcup\limits_{k=0}^{\infty}}
T\left[  P^{M,I},I\right]  ^{k}\left(  \emptyset\right)  =M.$

\emph{Answer sets} are defined in analogy to stable models, but taking into
account that atoms may be preceded by classical negation and that
$(a,\mathbf{p)}$ and $\left(  -a,\mathbf{p}\right)  $ are mutually exclusive
in answer sets.

\section{The Splitting Set Theorem for Hybrid ASP}

We will now introduce additional notation that will be used
throughout the rest of the paper.

Without loss of generality assume that all advancing rules are of the form%
\[
a:-B_{1};\;...;\;B_{n}:O,A
\]
and all of stationary rules are of the form%
\[
a:-B_{1};\;...;\;B_{n}:O,H
\]
where $a$ is a literal, $B_{1}$, ..., $B_{n}$ are blocks, $O$ is a constraint
set, $A$ is an advancing algorithm, and $H$ is a boolean algorithm.

Let $M$ be a set of literals and generalized position pairs, and let
$\mathbf{p}$ be a generalized position. Define%
\[
M|_{\mathbf{p}}\equiv\{\left(  a,\mathbf{q}\right)  \in M\;:\;\mathbf{q=p}\}
\]%
\[
At\left(  M\right)  \equiv\{a\;:\;\left(  a,\mathbf{p}\right)  \in M\}
\]

Let $U\subseteq Lit_{At}\times S$. We say that $U$ is a \emph{splitting set of
}$P$ \emph{with initial condition} (w.i.c.) $J$ if for all $r\in P$

\begin{enumerate}
\item if $r$ is advancing and $(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS\left(  r\right)  $ and $\mathbf{p}\in Adv\left(  r\right)  \left(
\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}\right)  $ and $\left(  a,\mathbf{p}%
\right)  \in U$ then both for $i=1,...,n$, $B_{i}\times\mathbf{p}_{i}\subseteq
U$ and $\{\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}\}\subseteq GP_{J}\left(
U\right)  $.

\item if $r$ is stationary and $(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS\left(  r\right)  $ and $\left(  a,\mathbf{p}_{n}\right)  \in U$ then both
for $i=1,...,n$, $B_{i}\times\mathbf{p}_{i}\subseteq U$ and $\{\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n}\}\subseteq GP_{J}\left(  U\right)  $.
\end{enumerate}

\bigskip As in the case of the original splitting set theorem
\cite{DBLP:conf/iclp/LifschitzT94} the splitting set $U$ acts to split Hybrid
ASP\ program $P$ into the part that can derive $U$ or one of its subsets, and
the remaining part of $P$, which can derive $At\times S\backslash U$ or one of
its subsets. The difference, however, is that for a given rule the conclusion
of the rule may be in $U$ for some n-tuples of generalized positions
($\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}$) and not for others. So, the
splitting set splits not only the program, but the rules themselves. This will
be elaborated below.

As in the case of the original splitting set theorem we identify by
$b_{U}\left(  P\right)  $ a set of new rules that capture the rules and
generalized positions that may contribute to generating $U$.

Define $Rules_{b}\left(  U,P\right)  $ as%
\[
\{\;r\in P\;:\;\text{if }r\text{ is advancing and there exists }%
(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)
\]%
\[
\text{and }\mathbf{p}\in Adv\left(  r\right)  (\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\text{ such that }\left(  a,\mathbf{p}\right)  \in U
\]%
\[
\text{if }r\text{ is stationary and there exists }(\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  \cap Bool\left(  r\right)
\]%
\[
\text{such that }\left(  a,\mathbf{p}_{n}\right)  \in U\;\}
\]

In other words, $Rules_{b}\left(  U,P\right)  $ is the set of all rules of $P$
that could contribute to $U$ for some tuple of generalized positions.

For an advancing rule $r$ let%
\[
CS_{b}\left(  U,r\right)  \equiv\{\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS\left(  r\right)  \;:
\]%
\[
\text{there exists }p\in Adv\left(  r\right)  (\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\text{ such that }\left(  a,\mathbf{p}\right)  \in
U\;\}
\]

For a stationary rule $r$ let%
\[
CS_{b}\left(  U,r\right)  \equiv\{\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS\left(  r\right)  \cap Bool\left(  r\right)  \;:\;\left(  a,\mathbf{p}%
_{n}\right)  \in U\;\}
\]

That is $CS_{b}\left(  U,r\right)  $ are all the generalized position tuples
for which $r$ could contribute to $U$.

For an advancing rule $r\in Rules_{b}\left(  U,P\right)  $ define
$Adv_{b}\left(  U,r\right)  $ by%
\[
Adv_{b}\left(  U,r\right)  (\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}%
)\equiv\{\;\mathbf{p}\;:\ \mathbf{p}\in Adv\left(  r\right)  (\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})
\]%
\[
\text{such that }\left(  a,\mathbf{p}\right)  \in U\text{ if }(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})\in CS_{b}\left(  U,r\right)  \;\}
\]

$Adv_{b}\left(  U,r\right)  $ is an advancing algorithm that for any tuple of
generalized positions will only generate those $\mathbf{p}$ that contribute to
$U$.

For an advancing rule $r$ let%
\[
b_{U}\left(  r\right)  \equiv head\left(  r\right)  :-\;body\left(  r\right)
:CS_{b}\left(  U,r\right)  ,\;Adv_{b}\left(  U,r\right)
\]

For a stationary rule $r$ let%
\[
b_{U}\left(  r\right)  \equiv head\left(  r\right)  :-\;body\left(  r\right)
:CS_{b}\left(  U,r\right)  ,\;Bool\left(  r\right)
\]

Define the \emph{bottom of }$P$ with respect to $U$, $b_{U}\left(  P\right)  $
as%
\[
b_{U}\left(  P\right)  \equiv\{\;b_{U}\left(  r\right)  \;:\;r\in
Rules_{b}\left(  U,P\right)  \;\}
\]
The idea is that just like in \cite{DBLP:conf/iclp/LifschitzT94},
$b_{U}\left(  P\right)  $ forms only those rules that could contribute to $U$,
and so $X$ will be an answer set of $b_{U}\left(  P\right)  $ w.i.c. $J$ iff
$M\cap U=X$ for some answer set $M$ of $P$ w.i.c. $J$.

\bigskip

We will now proceed to define $\epsilon_{U}\left(  P,X\right)  $ with the
understanding that the same rule may contribute to $U$ for some generalized
position tuples and contribute to $Lit_{At}\times S\;\backslash U$ for others.

First, we need to identify remainder $\operatorname*{Rem}\left(  U,P\right)  $
of $Rules_{b}\left(  U,P\right)  $ not captured by $b_{U}\left(  P\right)  $.
That is we need to identify the parts contributing to the complement of $U$ of
those rules that have other parts contributing to $U$. This is due to an
important difference between Hybrid ASP\ and ASP. In ASP a rule contributes a
single conclusion. Thus if ASP rule contributes to the splitting set then it
must be in the bottom of the program. In Hybrid ASP, however, a rule acts more
like a collection of rules contributing different conclusions for different
generalized position tuples. Consequently, the parts of the rules that
contribute to the complement of the splitting set need to be separated from
those that contribute to the splitting set itself. We will now proceed with
the definition.

For an advancing rule $r$ define%
\[
CS_{\operatorname*{Rem}}\left(  U,r\right)  \equiv\{\ (\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  :
\]%
\[
\text{there exists }\mathbf{p}\in Adv\left(  r\right)  (\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})\text{ }\left(  a,\mathbf{p}\right)  \notin U\;\}
\]

For a stationary $r$ define%
\[
CS_{\operatorname*{Rem}}\left(  U,r\right)  \equiv\{\;(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  \cap Bool\left(  r\right)
:\;\left(  a,\mathbf{p}_{n}\right)  \notin U\;\}
\]

That is, $CS_{\operatorname*{Rem}}\left(  U,r\right)  $ contains those
generalized position tuples such that for them the rule $r$ contributes to the
complement of $U$.

For an advancing rule $r\in Rules_{b}\left(  U,P\right)  $ and $(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})$ define
\[
Adv_{\operatorname*{Rem}}(U,r)(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}%
)\equiv\left\{
\begin{array}
[c]{l}%
\{\mathbf{p}\;:\;\mathbf{p}\in Adv\left(  r\right)  (\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\text{ s.t. }\left(  a,\mathbf{p}\right)  \notin
U\;\}\\
\text{ \ \ \ if }(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS_{\operatorname*{Rem}}\left(  U,r\right)  \\
\emptyset\text{ if }(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\notin
CS_{\operatorname*{Rem}}(U,r)
\end{array}
\right.
\]

\bigskip That is $Adv_{\operatorname*{Rem}}(U,r)$ is a restriction of
$Adv\left(  r\right)  $ to those generalized positions such that for them $r$
contributes to the complement of $U$.

When $CS_{\operatorname*{Rem}}\left(  U,r\right)  \neq\emptyset$ define
\[
\operatorname*{Rem}\left(  U,r\right)  \equiv\left\{
\begin{array}
[c]{c}%
head\left(  r\right)  :-\;body\left(  r\right)  :CS_{\operatorname*{Rem}%
}\left(  U,r\right)  ,\;Adv_{\operatorname*{Rem}}\left(  U,r\right)  \text{ if
}r\text{ is advancing}\\
head\left(  r\right)  :-\;body\left(  r\right)  :CS_{\operatorname*{Rem}%
}\left(  U,r\right)  ,\;Bool\left(  r\right)  \text{ if }r\text{ is
stationary}%
\end{array}
\right.
\]

\bigskip In other words, $\operatorname*{Rem}\left(  U,r\right)  $ is the part
of $r$ that contributes to the complement of $U$.

Define%
\[
\operatorname*{Rem}\left(  U,P\right)  \equiv\{\;\operatorname*{Rem}\left(
U,r\right)  :\;r\in Rules_{b}\left(  U,P\right)  \text{ and }%
CS_{\operatorname*{Rem}}\left(  U,r\right)  \neq\emptyset\;\}
\]

\bigskip That is $\operatorname*{Rem}\left(  U,P\right)  $ contain those parts
of the rules in $Rules_{b}\left(  U,P\right)  $ that contribute to the
complement of $U$.

Let $X\subseteq U$. For a rule $r$ define
\[
CS_{\epsilon}\left(  U,r,X\right)  \equiv\{\;(\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  :
\]%
\[
\text{for }i=1,\;...,\;n\text{ }\{B_{i}^{+}\times\mathbf{p}_{i}\}\cap
U\subseteq X\text{ and }\{B_{i}^{-}\times\mathbf{p}_{i}\}\cap X=\emptyset\;\}
\]

That is $CS_{\epsilon}\left(  U,r,X\right)  $ is the set of those generalized
position tuples such that for them the "projection"\ of $body\left(  r\right)
$ onto $U$ is satisfied by $X$.

Finally
\[
\epsilon_{U}\left(  P,X\right)  \equiv\{
\]%
\[
r^{\prime}\equiv a:-\;B_{1}\backslash At\left(  U|_{\mathbf{p}_{1}}\right)
;\;...;\;B_{n}\backslash At\left(  U|_{\mathbf{p}_{n}}\right)
:\;\{(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\},\;Q\text{ }|
\]%
\[
r\equiv a:-\;B_{1};\;...;\;B_{n}:O,Q\;\in\;\{\;r\in P:\;CS\left(  \epsilon
_{U},r,X\right)  \neq\emptyset\;\}\text{ and}%
\]%
\[
(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in CS_{\epsilon}\left(  U,r,X\right)
\;\}
\]

In other words, for every rule $r\in P$ such that$\;CS\left(  \epsilon
_{U},r,X\right)  \neq\emptyset$ and for every $(\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\in CS_{\epsilon}\left(  U,r,X\right)  $ where the
"projection" of $body\left(  r\right)  $ onto $U$ is satisfied by $X$ at
$(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})$, we add to $\epsilon_{U}\left(
P,X\right)  $ a rule $r^{\prime}$, which is a part of rule $r$ that will be
active only for that $(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})$ with the
"projection"\ part removed.

\bigskip

\textbf{Theorem 1. }\emph{(The Splitting Set Theorem for Hybrid ASP). Let }%
$P$\emph{ be a Hybrid ASP\ program over }$Lit_{At}\times S$\emph{. Let
}$U\subseteq Lit_{At}\times S$\emph{ be a splitting set of }$P$\emph{ w.i.c.
}$J\subseteq S$\emph{. A set }$M$\emph{ is a answer set of }$P$\emph{ w.i.c.
}$J$\emph{ iff }$X\equiv M\cap U$\emph{ is a answer set of }$b_{U}\left(
P\right)  $\emph{ w.i.c. }$J$\emph{ and }$M\backslash U$\emph{ is a answer set
of }$\epsilon_{U}(P\backslash Rules_{b}\left(  U,P\right)  \cup
\operatorname*{Rem}\left(  U,P\right)  ,\;X)$\emph{ w.i.c. }$GP_{J}\left(
X\right)  $\emph{.}

\bigskip

\emph{Sketch of a proof. }We first prove that if $M$ is an answer set of $P$
w.i.c. $J$ then $X\equiv M\cap U$ is an answer set of $b_{U}\left(  P\right)
$ w.i.c. $J$. That is, we want to show that $X=%
{\displaystyle\bigcup\limits_{k=0}^{\infty}}
T\left[  b_{U}\left(  P\right)  ^{X,J},J\right]  ^{k}\left(  \emptyset\right)
$. In $\supseteq$ direction we show by induction on $k$ in one-step
provability operator $T\left[  b_{U}\left(  P\right)  ^{X,J},J\right]  ^{k}$
that if a rule $b_{U}\left(  r\right)  ^{X,J}$ in $b_{U}\left(  P\right)
^{X,J}$ derives $\left(  a,\mathbf{p}\right)  $ in $T\left[  b_{U}\left(
P\right)  ^{X,J},J\right]  ^{k+1}\left(  \emptyset\right)  $, then the rule
$r^{M,J}$ must derive $\left(  a,\mathbf{p}\right)  $ in $T\left[
P^{M,J},J\right]  ^{m+1}\left(  \emptyset\right)  $ for some $m$. In
$\subseteq$ direction we show by induction on $k$ in $T\left[  P^{M,J}%
,J\right]  ^{k}\left(  \emptyset\right)  $ that if $r^{M,J}$ derives $\left(
a,\mathbf{p}\right)  $ in $T\left[  P^{M,J},J\right]  ^{k+1}\left(
\emptyset\right)  $ where $\left(  a,\mathbf{p}\right)  \in U$, then
$b_{U}\left(  r\right)  ^{X,J}$ derives $\left(  a,\mathbf{p}\right)  $ in
$T\left[  b_{U}\left(  P\right)  ^{X,J},J\right]  ^{m+1}\left(  \emptyset
\right)  $ for some $m$.

We then proceed to prove that if $M$ is an answer set of $P$ w.i.c. $J$, and
$Y\equiv M\backslash U$ then $Y$ is an answer set of $Q\equiv\epsilon
_{U}(P\backslash Rules_{b}\left(  U,P\right)  \cup\operatorname*{Rem}\left(
U,P\right)  ,X)$\emph{ w.i.c. }$L\equiv GP_{J}\left(  X\right)  $. That is, we
want to show that $Y=%
{\displaystyle\bigcup\limits_{k=0}^{\infty}}
T\left[  Q^{Y,L},L\right]  ^{k}\left(  \emptyset\right)  $. In $\supseteq$
direction we prove by induction that if $r^{Y,L}$ derives $\left(
a,\mathbf{p}\right)  $ in $T\left[  Q^{Y,L},L\right]  ^{k+1}\left(
\emptyset\right)  $ then there is a corresponding rule $q^{M,J}$ in $P^{M,J}$
that derives $\left(  a,\mathbf{p}\right)  $ in $T\left[  P^{M,J},J\right]
^{m+1}\left(  \emptyset\right)  $ for some $m$. In $\subseteq$ direction we
prove by induction on $k$ in $T\left[  P^{M,J},J\right]  ^{k}\left(
\emptyset\right)  $ that if $q^{M,J}$ derives $\left(  a,\mathbf{p}\right)  $
in $T\left[  P^{M,J},J\right]  ^{k+1}\left(  \emptyset\right)  $ where
$\left(  a,\mathbf{p}\right)  \in M\backslash U$ then there is a corresponding
$r$ in $Q^{Y,L}$ that derives $\left(  a,\mathbf{p}\right)  $ in $T\left[
Q^{Y,L},L\right]  ^{m+1}\left(  \emptyset\right)  $ for some $m$.

To finish the proof we need to show that if $X\subseteq U$ is an answer set of
$b_{U}\left(  P\right)  $ w.i.c. $J$ and $Y\subseteq U^{C}$ is an answer set
of $Q$ w.i.c. $L$ then $M\equiv X\cup Y$ is an answer set of $P$ w.i.c. $J$.
That is we want to show that $M=%
{\displaystyle\bigcup\limits_{k=0}^{\infty}}
T\left[  P^{M,J},J\right]  ^{k}\left(  \emptyset\right)  $. We do so by
induction in both directions in a manner similar to the previous part of the
proof. $\square$

\bigskip

Similar to the Splitting Sequence Theorem of
\cite{DBLP:conf/iclp/LifschitzT94} we also prove the Splitting Sequence
Theorem for Hybrid ASP.

\textbf{Theorem 2.}\emph{ (The Splitting Sequence Theorem for Hybrid ASP). Let
}$\left\langle U_{\alpha}\right\rangle _{\alpha<\mu}$ \emph{be a monotone
continuous sequence of splitting sets for a Hybrid ASP\ program }$P$\emph{
over }$At\times S$\emph{ w.i.c. }$J\subseteq S$\emph{, and }$%
{\displaystyle\bigcup\limits_{\alpha<\mu}}
U_{\alpha}=Lit_{At}\times S$.\emph{ }$M$\emph{ is an answer set of }$P$\emph{
w.i.c. }$J$\emph{ iff }$M=%
{\displaystyle\bigcup\limits_{\alpha<\mu}}
X_{\alpha}$\emph{ for a sequence }$\left\langle X_{\alpha}\right\rangle
_{\alpha<\mu}$\emph{ s.t.}

\begin{itemize}
\item $X_{0}$\emph{ is an answer set of }$b_{U_{0}}\left(  P\right)  $\emph{
w.i.c. }$J$

\item \emph{for any }$\alpha$\emph{ such that }$\alpha+1<\mu$\emph{
}$X_{\alpha+1}$\emph{ is an answer set for }

$\epsilon_{U_{\alpha}}(b_{U_{\alpha+1}}\left(  P\right)  \;\backslash
\;Rules_{b}(U_{\alpha},b_{U_{\alpha+1}}\left(  P\right)  )\cup
\operatorname*{Rem}(U_{\alpha},b_{U_{\alpha+1}}\left(  P\right)  ),\;%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta})$\emph{ w.i.c.}\emph{ }$L_{\alpha}\equiv GP_{J}(%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta})$\emph{ and }$X_{\alpha+1}=M\cap(U_{\alpha+1}\backslash U_{\alpha}%
)$\emph{ and }$%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta}$\emph{ is an answer set of }$b_{U_{\alpha}}\left(  P\right)  $\emph{
w.i.c. }$J$\emph{.}
\end{itemize}

\bigskip

The proof proceeds by the induction on $\alpha$ and is a direct application of
The Splitting Set Theorem for Hybrid ASP.

In the Splitting Sequence Theorem for Hybrid ASP, $b_{U_{\alpha+1}}\left(
P\right)  $ is a program that derives $%
{\displaystyle\bigcup\limits_{\beta\leq\alpha+1}}
X_{\beta}$ as its answer set w.i.c. \ $J$. Now, $%
{\displaystyle\bigcup\limits_{\beta\leq\alpha+1}}
X_{\beta}$ $\subseteq%
{\displaystyle\bigcup\limits_{\beta\leq\alpha+1}}
U_{\beta}$. So, to derive $X_{\alpha+1}$ (i.e. the subset of $%
{\displaystyle\bigcup\limits_{\beta\leq\alpha+1}}
X_{\beta}$ that is in $U_{\alpha+1}\backslash U_{\alpha}$) we need to remove
from $b_{U_{\alpha+1}}\left(  P\right)  $ the rules that derive $%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta}$. That is accomplished by subtracting from $b_{U_{\alpha+1}}\left(
P\right)  $ the rules $Rules_{b}(U_{\alpha},b_{U_{\alpha+1}}\left(  P\right)
)$. Nevertheless, this subtracts too much as some of the rules in
$Rules_{b}(U_{\alpha},b_{U_{\alpha+1}}\left(  P\right)  )$ contribute to
$X_{\alpha+1}$ for some generalized position tuples. The parts of those rules
that contribute to $X_{\alpha+1}$ are $\operatorname*{Rem}(U_{\alpha
},b_{U_{\alpha+1}}\left(  P\right)  )$, which we then add back. Applying
$\epsilon_{U_{\alpha}}$ operator to the resulting program (i.e. $b_{U_{\alpha
+1}}\left(  P\right)  \;\backslash\;Rules_{b}(U_{\alpha},b_{U_{\alpha+1}%
}\left(  P\right)  )\cup\operatorname*{Rem}(U_{\alpha},b_{U_{\alpha+1}}\left(
P\right)  )$) then removes the "useless"\ part of the rules with respect to $%
{\displaystyle\bigcup\limits_{\beta\leq\alpha}}
X_{\beta}$.

\bigskip

\section{An Application:\ Computing Answer Sets of Hybrid ASP\ Programs}

One of the applications of the Splitting Sequence Theorem for Hybrid ASP\ is
proving the correctness of a certain algorithm for computing answer sets of
certain types of Hybrid ASP\ programs. We will consider only the programs
where the set of generalized positions $S$ is such that if $\mathbf{p}\in S$
then $t\left(  \mathbf{p}\right)  =k\cdot\Delta t$ where $k\in%
\mathbb{N}
$, and for any advancing rule $r$ of any arity $n$, for any $(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})\in S^{n}$ we have that for all $\mathbf{q\in
}Adv\left(  r\right)  (\mathbf{p}_{1},...,\;\mathbf{p}_{n})$, $t\left(
\mathbf{q}\right)  =t\left(  \mathbf{p}_{n}\right)  +\Delta t$. That is, these
are the programs with generalized positions with discrete times of the form
$k\Delta t$, and whenever an advancing algorithm produces a new generalized
position, that generalized position has time larger by $\Delta t$ than the
largest time in the input arguments. All applications of Hybrid ASP known to
the author\ are restricted to such programs. This is the case for using Hybrid
ASP\ to diagnose failure of data processing pipelines, as described in
\cite{BomansonB19} and \cite{BrikX20}. It is the case for the Hybrid
ASP\ programs that are the result of translation from action languages Hybrid
AL \cite{BrikR17} and Hybrid ALE \cite{BomansonB19}. It is also the case for
using Hybrid ASP to compute optimal finite horizon policies in dynamic domains
\cite{HybridASP2}.

\textbf{The algorithm.}

We will first describe the algorithm informally. We will use some of the new
notation which will be defined further below. The algorithm is based on the
observation that in Hybrid ASP the facts in the "future"\ cannot affect the
facts in the "past". That is for any two generalized position $\mathbf{p}$ and
$\mathbf{q}$, if $t\left(  \mathbf{p}\right)  <t\left(  \mathbf{q}\right)  $
then the state at $\mathbf{q}$ cannot be used to derive the state at
$\mathbf{p}$ (but the state at $\mathbf{p}$ can be used to derive the state at
$\mathbf{q}$). Consequently, it should be possible to first derive the states
at some minimal time $t_{\min}$, then derive the states at the time $t_{\min
}+\Delta t$, then derive the states at time $t_{\min}+2\Delta t$ and so on.

Without the loss of generality, we will assume that for any initial condition
$J\subseteq S$, there exists $\mathbf{p}\in J$ such that $t\left(
\mathbf{p}\right)  =0$. Let $P$ be a Hybrid ASP\ program over $Lit_{At}\times
S$. Let $J\subseteq S$ be an initial condition. The algorithm will be defined
inductively. Suppose the set $N$ of all the (literal, generalized
position)\ pairs for the generalized positions with time up to $k\cdot\Delta
t$ is derived by the algorithm for some $k$. The algorithm will first identify
all the advancing rules $Rules_{Adv}\left(  P,N,k\Delta t\right)  $ that could
derive generalized positions with time $\left(  k+1\right)  \cdot\Delta t$.
These are the advancing rules $r$ such that $N$ satisfies their body for some
$(\mathbf{p}_{1},...,\;\mathbf{p}_{n})\in CS\left(  r\right)  $, where
$n=arity\left(  r\right)  $ and the time of $\mathbf{p}_{n}$ is $k\cdot\Delta
t$. The set of the "next" generalized positions (i.e. the set of generalized
positions with time $\left(  k+1\right)  \cdot\Delta t$) is derived by
choosing a subset of the set of all the generalized positions derived by these
rules. To formally define such a choice of a subset we introduce a concept of
an \emph{advancing selector} $F$, which is a function s.t. for $M\subseteq
Lit_{At}\times S$ and $Z\subseteq S$, $F\left(  M,Z\right)  $ is a subset of
$Z$. We will denote the set of "next" generalized positions derived in this
manner by $NextGP\left(  P,F,N,k\Delta t\right)  $.

Now, for every "next" generalized position $\mathbf{q}$ in $NextGP\left(
P,F,N,k\Delta t\right)  $ derived by an advancing rule $r\in Rules_{Adv}%
\left(  P,N,k\Delta t\right)  $, it must be that $(head\left(  r\right)
,\;\mathbf{q})$ is derived. So, for every $\mathbf{q}$ there is a set of
literals that will be derived at $\mathbf{q}$ by the advancing rules in
$Rules_{Adv}\left(  P,N,k\Delta t\right)  $. This set of literals will be
denoted by $Head_{Adv}(P,N,\mathbf{q})$.

Next we turn our attention to the role of the stationary rules in deriving
hybrid state at a "next" generalized position $\mathbf{q}$. There is a set of
stationary rules that can contribute to the hybrid state at $\mathbf{q}$. If
such a stationary rule $r$ has $n$ blocks, then the first $n-1$ blocks are
satisfied by $N$ (at some generalized positions $\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n-1}$) and $(\mathbf{p}_{1},\;...,\;\mathbf{p}%
_{n-1},\;\mathbf{q})$ are in $CS\left(  r\right)  \cap Bool\left(  r\right)
$. Thus, only the last block, which we will denote by $B_{n}$ needs to be
evaluated at $\mathbf{q}$. Thus, the relevant part of such a stationary rule
$r$ is a regular ASP rule of the form $head\left(  r\right)  :-B_{n}$. All
such regular ASP\ rules applicable at $\mathbf{q}$ will be denoted by
$\operatorname*{Red}\nolimits_{App}(P,\ N,\;\mathbf{q})$. A state at
$\mathbf{q}$ is then an answer set of a regular ASP\ program
$\operatorname*{Red}\nolimits_{App}(P,\ N,\;\mathbf{q})\cup\{[h:-]\;:\;h\in
Head_{Adv}(P,N,\mathbf{q})\}$. To formally define such a choice we will use a
concept of a \emph{stationary selector }$D$, which we will define further below.

We will now define the algorithm formally.

For a set $N\subseteq Lit_{At}\times S$ and generalized positions $\mathbf{p}$
and $\mathbf{q}$, let
\[
Rules_{Adv}\left(  P,N,k\Delta t\right)  \equiv\{r\in P\;:\;r\text{ is an
advancing rule and there is }%
\]%
\[
\text{ }(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in GP_{J}\left(  N\right)
^{n}\cap CS\left(  r\right)  \text{ with }t\left(  \mathbf{p}_{n}\right)
=k\cdot\Delta t\text{ and }N\models_{J}(body\left(  r\right)  ,\;(\mathbf{p}%
_{1},...,\;\mathbf{p}_{n})\}
\]

Let $\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}\in GP_{J}\left(  N\right)  $. We
define the set of advancing rules active at $\mathbf{p}_{1},\;...,\;\mathbf{p}%
_{n}$ relative to $N$ as%
\[
Rules_{Adv}(P,\;N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))\equiv\{r\in
Rules_{Adv}\left(  P,\;N,\;t\left(  \mathbf{p}_{n}\right)  \right)  \;:\text{
}(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  \}.
\]
That is, $Rules_{Adv}(P,\;N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))$ is the
set of the advancing rules whose body is satisfied by $N$ at $(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n})$ and $(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in
CS\left(  r\right)  $.

We define the set of "next" generalized positions at $\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n}$ relative to $N$ as%
\[
NextGP(P,N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))\equiv%
{\displaystyle\bigcup\limits_{r\in Rules_{Adv}(P,\;N,\;(\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n}))}}
Adv\left(  r\right)  (\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}).
\]
That is $NextGP(P,N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))$ is the set of
"next" generalized positions generated by any advancing rule active at
$\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}$ relative to $N$.

For a time $k\cdot\Delta t$, we define the set of all the "next" generalized
positions relative to $N$, $k\cdot\Delta t$ and an advancing selector $F$ as%
\[
NextGP(P,F,N,k\Delta t)\equiv F(N,\;%
{\displaystyle\bigcup\limits_{\substack{_{\substack{n\geq1\\\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n}\in GP_{J}\left(  N\right)  }}\\t\left(
\mathbf{p}_{n}\right)  =k\Delta t}}}
NextGP(P,N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))\;).
\]

The set of all heads at $\mathbf{q}\in NextGP\left(  P,F,N,k\Delta t\right)  $
relative to $N$ is then%
\[
Head_{Adv}(P,\;N,\;\mathbf{q})\equiv\{\;head\left(  r\right)  \;:\text{there
exists }\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}\in GP_{J}\left(  N\right)
\text{ and}%
\]%
\[
\text{ }r\in Rules_{Adv}(P,\;N,\;(\mathbf{p}_{1},\;...,\mathbf{\;p}_{n}))
\]%
\[
\text{such that }\mathbf{q}\in Adv\left(  r\right)  (\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\}\text{.}%
\]

Let $\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}\in GP_{J}\left(  N\right)  $. We
define the set of stationary rules active at $\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n}$ relative to $N$ as%
\[
Rules_{Stat}(P,\;N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))\equiv\{r\in
P\;:r\text{ is stationary and}%
\]%
\[
\text{ }(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  \cap
Bool\left(  r\right)  \text{ and for }i=1,\;...,\;n-1\text{ }N\models
_{J}(B_{i},\;\mathbf{p}_{i})\;\}\text{.}%
\]
That is $Rules_{Stat}(P,\;N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n}))$ is the
set of stationary rules with $n-1$ blocks satisfied by $N$ at $\mathbf{p}%
_{1},\;...,\;\mathbf{p}_{n-1}$ respectively, and $(\mathbf{p}_{1}%
,\;...,\;\mathbf{p}_{n})\in CS\left(  r\right)  \cap Bool\left(  r\right)  $.

We define a \emph{stationary selector }$D$ to be a function such that for
$M\subseteq At\times S$ for $\mathbf{z}\in S$ for an ASP\ program $U$,
$D(M,\;\mathbf{z},\;U)$ is an answer set of $U$. That is, a stationary
selector chooses one of answer sets of a regular ASP\ programs $U$.

For a stationary rule $r$ of the form $a:-B_{1};\;...;\;B_{n}:O,H\,$, we
define an \emph{applicable reduct} of $r$
\[
\operatorname*{Red}\nolimits_{App}\left(  r\right)  \equiv\{a:-\;B_{n}%
\}\text{.}%
\]

For $\mathbf{z}\in NewGP(P,F,N,k\Delta t)$ we define the active reduct of $P$
at $\mathbf{z}$ relative to $N$ as%
\[
\operatorname*{Red}\nolimits_{App}(P,\;N,\;\mathbf{z})\equiv
\{\operatorname*{Red}\nolimits_{App}\left(  r\right)  :\;\text{there exists
}n\geq1\text{ and }(\mathbf{p}_{1},\;...,\;\mathbf{p}_{n-1})\in GP_{J}\left(
N\right)  ^{n-1}%
\]%
\[
\text{ such that }r\in Rules_{Stat}(P,\;N,\;(\mathbf{p}_{1},\;...,\;\mathbf{p}%
_{n-1},\;\mathbf{z})\;\}
\]
\bigskip

Finally, for $N\subseteq At\times S$ and $i\in%
\mathbb{N}
$ let $N\left[  i\right]  \equiv\{(a,\;\mathbf{p})\in N:\;t\left(
\mathbf{p}\right)  =i\cdot\Delta t\}$. Similarly for $Z\subseteq S$,
$Z[i]\equiv\{\mathbf{p}\in Z\;:\;t\left(  \mathbf{p}\right)  =i\cdot\Delta
t\}$.

We are now ready to formally specify our algorithm. We define a sequence of
sets $\left\langle Y_{i}\right\rangle _{i\geq0}$ , $Y_{i}\subseteq(At\times
S)[i]$ as follows:%
\[
Y_{0}\equiv%
{\displaystyle\bigcup\limits_{\mathbf{z}\in J\left[  0\right]  }}
D(\emptyset,\;\mathbf{z},\;\operatorname*{Red}\nolimits_{App}(P,\;\emptyset
,\;\mathbf{z))\times z}%
\]

That is, the state at any generalized position $\mathbf{z}\in J$ with time
equal to $0$ is determined by taking all the stationary rules $r$ with one
block (i.e. rules of the form $a:-B:O,H$ ) such that $\mathbf{z}\in O\cap H$,
composing a regular ASP\ program from the reducts of the form $a:-B$ derived
from those rules, and then finding an answer set of that program.

Now, suppose $Y_{i}$ are defined for $0\leq i\leq k$ and $Y_{k}\neq\emptyset$.
Let%
\[
Z_{k+1}\equiv NextGP(P,F,\;%
{\displaystyle\bigcup\limits_{i=0}^{k}}
Y_{i},\;k\Delta t).
\]
That is $Z_{k+1}$ is the set of generalized positions with time $\left(
k+1\right)  \Delta t$ derived by the advancing rules $Rules_{Adv}\left(  P,\;%
{\displaystyle\bigcup\limits_{i=0}^{k}}
Y_{i},\;k\Delta t\right)  $.

Let%
\[
Y_{k+1}\equiv%
{\displaystyle\bigcup\limits_{\mathbf{z}\in Z_{k+1}}}
D(%
{\displaystyle\bigcup\limits_{i=0}^{k}}
Y_{i},\;\mathbf{z},\;\operatorname*{Red}\nolimits_{App}(P,\;%
{\displaystyle\bigcup\limits_{i=0}^{k}}
Y_{i},\;\mathbf{z})\cup
\]%
\[
\{[a:-]\;:\;a\in Head_{Adv}(P,\;%
{\displaystyle\bigcup\limits_{i=0}^{k}}
Y_{i},\;\mathbf{z})\})\;\times\;\mathbf{z}%
\]
if $D(...)\neq\emptyset$ and $Y_{k+1}\equiv\emptyset$ otherwise.

That is, $Y_{k+1}$ is a collection of hybrid states $\left(  Y_{k+1}%
|_{\mathbf{z}},\mathbf{z}\right)  $ where $\mathbf{z\in Z}_{k+1}$, and where
$Y_{k+1}|_{\mathbf{z}}$ is an answer set of a regular ASP\ program composed of
the active reducts of the stationary rules that can contribute to $\mathbf{z}$
and the heads of the advancing rules that derive $\mathbf{z}$.\newline

\textbf{Theorem 3.} $M$ is an answer set of $P$ w.i.c. $J$ iff there is
advancing selector $F$ and a stationary selector $D$ such that $%
{\displaystyle\bigcup\limits_{i=0}^{\infty}}
Y_{i}=M$ with $F$ and $D$.

\bigskip

\emph{Sketch of a proof.} We begin by specifying a sequence of splitting sets
$\left\langle U_{i}\right\rangle _{i=0}^{\infty}$ defined as%
\[
U_{i}=Lit_{At}\times\{\mathbf{p}:\;\mathbf{p}\in S\text{ and }0\leq t\left(
\mathbf{p}\right)  \leq i\Delta t\;\}
\]

We then first show that $Y_{0}$ is an answer set of $b_{U_{0}}\left(
P\right)  $ w.i.c. $J$. The rules that can contribute to $Y_{0}$ are
stationary-1 rules $r$ such as $CS\left(  r\right)  \cap Bool\left(  r\right)
\cap J\left[  0\right]  \neq\emptyset$. These rules will contribute regular
ASP\ rules to $\operatorname*{Red}\nolimits_{App}(P,\;\emptyset,\;\mathbf{z})$
for every $\mathbf{z}\in J\left[  0\right]  $. We then show that
$D(\emptyset,\;\mathbf{z},\;\operatorname*{Red}\nolimits_{App}(P,\emptyset
,\mathbf{z}))$ is an answer set of $\operatorname*{Red}\nolimits_{App}%
(P,\emptyset,\mathbf{z})$ iff $D(\emptyset,\;\mathbf{z},\;\operatorname*{Red}%
\nolimits_{App}(P,\emptyset,\mathbf{z}))\times\mathbf{z}$ is an answer set of
$b_{U_{0}}\left(  P\right)  $ w.i.c. $J$.

The rest is proven by induction using The Splitting Sequence Theorem. That is
$M\left[  k+1\right]  $ is an answer set of $E=\epsilon_{U_{k}}(b_{U_{k+1}%
}\left(  P\right)  \backslash Rules_{b}(U_{k},b_{U_{k+1}}\left(  P\right)
)\cup\operatorname*{Rem}(U_{k},b_{U_{k+1}}\left(  P\right)  ),\;%
{\displaystyle\bigcup\limits_{i\leq k}}
M\left[  i\right]  )$ w.i.c. $GP_{J}(L)$, where $L=%
{\displaystyle\bigcup\limits_{i\leq k}}
M\left[  i\right]  $ iff there exists an advancing selector $F$ and a
stationary selector $D$ such that $M\left[  k+1\right]  $ is equal to
$Y_{k+1}$ as defined by the algorithm.

\bigskip For the forward direction of the inductive step we define
$F(N,\;Y)\equiv Y\cap GP\left(  M\right)  $. We define
\[
D(N,\;\mathbf{p},\;Q)\equiv\left\{
\begin{array}
[c]{c}%
At\left(  N|_{\mathbf{p}}\right)  \text{ if }At\left(  N|_{\mathbf{p}}\right)
\text{ is an answer set of }Q\\
\emptyset\text{ otherwise}%
\end{array}
\right.
\]

We then show $GP\left(  M\left[  k+1\right]  \right)  =NextGP(P,F,L,k\Delta
t)$. We then use the induction on one-step provability operator\ $T\left[
E^{M\left[  k+1\right]  ,\;GP_{J}\left(  L\right)  },GP_{J}\left(  L\right)
\right]  ^{j}$ to show that if $M\left[  k+1\right]  $ is an answer set of $E$
w.i.c. $GP_{J}\left(  L\right)  $ then $M\left[  k+1\right]  |_{\mathbf{p}%
}=Y_{k+1}|_{\mathbf{p}}$. That is we show that if $M\left[  k+1\right]  $ is
an answer set of $E$ w.i.c. $GP_{J}\left(  L\right)  $ then the algorithm
derives it as $Y_{k+1}$.

For the reverse direction we first show $\{(head\left(  r\right)
,\;\mathbf{p}):\;r\in Head_{Adv}(P,\;L,\;\mathbf{p}),\;\mathbf{p}\in GP\left(
Y_{k}\right)  \}\subseteq T\left[  E^{Y_{k+1},GP_{J}\left(  L\right)  }%
,GP_{J}\left(  L\right)  \right]  ^{1}\left(  \emptyset\right)  $. That is we
show that the literals of $Head_{Adv}(P,\;L,\;\mathbf{p})$ are also derived by
$E$ at $\mathbf{p}$. We then use induction on one step provability $T\left[
K^{At\left(  Y_{k+1}|_{\mathbf{p}}\right)  }\right]  ^{i}$, where
$\ K\equiv\operatorname*{Red}\nolimits_{App}(P,L,\mathbf{p})$ to show that for
all $\mathbf{p}\in GP\left(  Y_{k+1}\right)  $ it is the case that%
\ $%
{\displaystyle\bigcup\limits_{i\geq0}}
T\left[  K^{At\left(  Y_{k+1}|_{\mathbf{p}}\right)  }\right]  ^{i}\left(
\emptyset\right)  \times\mathbf{p}\subseteq%
{\displaystyle\bigcup\limits_{j\geq0}}
T\left[  E^{Y_{k+1},GP_{J}\left(  L\right)  },GP_{J}\left(  L\right)  \right]
^{j}\left(  \emptyset\right)  $, for some $j$. That is, we show that the
literals derived by the regular ASP\ program $\operatorname*{Red}%
\nolimits_{App}(P,L,\mathbf{p})$ are also derived by $E$ at $\mathbf{p}$. But
this merely shows that $Y\equiv%
{\displaystyle\bigcup\limits_{i=0}^{\infty}}
Y_{i}$ $\subseteq$ $%
{\displaystyle\bigcup\limits_{j\geq0}}
T[P^{Y,J},J]\left(  \emptyset\right)  $. We also need to show that $%
{\displaystyle\bigcup\limits_{j\geq0}}
T[P^{Y,J},J]\left(  \emptyset\right)  \subseteq Y$.

We do that by using induction on one step provability operator $T\left[
E^{Y_{k+1},GP_{J}\left(  L\right)  },GP_{J}\left(  L\right)  \right]  ^{j}$ to
show that for all $\mathbf{p}\in GP\left(  Y_{k+1}\right)  $ it is the case
that $%
{\displaystyle\bigcup\limits_{j\geq0}}
T\left[  E^{Y_{k+1},GP_{J}\left(  L\right)  },GP_{J}\left(  L\right)  \right]
^{j}\left(  \emptyset\right)  $ is a subset of $%
{\displaystyle\bigcup\limits_{i\geq0}}
T\left[  K^{At\left(  Y_{k+1}|_{\mathbf{p}}\right)  }\right]  ^{i}\left(
\emptyset\right)  \times\mathbf{p}$.

This completes the proof of the theorem. $\square$

\bigskip The algorithm computes an answer set of the Hybrid ASP program $P$
w.i.c. $J$ inductively, by computing a subset of the answer set at time $0$,
then at time $\Delta t$, and so on through time $k\Delta t$. Moreover, the
aglorithm reduces the process of computing an answer set of a Hybrid ASP
program to the repeated application of two processes:\ the process of
computing the set of "next"\ generalized positions, and the process of
computing an answer set of a regular ASP\ program derived from advancing and
stationary Hybrid ASP\ rules applicable at these "next" generalized positions.

It's worth noting that the algorithm is a more general form of The Local
Algorithm \cite{HybridASP2}, variation of which is also discussed in
\cite{BomansonB19}.

\section{Conclusion}

The paper presents The Splitting Set Theorem for Hybrid ASP, which is\ the
equivalent for Hybrid ASP\ of the Splitting Set Theorem
\cite{DBLP:conf/iclp/LifschitzT94}, and the Splitting Sequence Theorem for
Hybrid ASP (which is the equivalent for Hybrid ASP\ of The Splitting Sequence
Theorem). The original Splitting Set Theorem proved to be a widely used
result. It is the author's hope that the new theorem will likewise prove to
have many applications. The paper discusses one of the applications of the
theorems to computing answer sets of Hybrid ASP\ programs.\newline%
\newline\textbf{Acknowledgements.} The author would like to thank Jori
Bomanson for insightful comments that helped to enhance the paper.

\bibliographystyle{eptcs}


text:
import { moduleForComponent, test } from 'ember-qunit';
import hbs from 'htmlbars-inline-precompile';

moduleForComponent('search/filter-rail', 'Integration | Component | search/filter rail', {
  integration: true
});

test('it renders', function(assert) {

  // Set any properties with this.set('myProperty', 'value');
  // Handle any actions with this.on('myAction', function(val) { ... });

  this.render(hbs`{{search/filter-rail}}`);

  assert.equal(this.$().text().trim(), '');

  // Template block usage:
  this.render(hbs`
    {{#search/filter-rail}}
      template block text
    {{/search/filter-rail}}
  `);

  assert.equal(this.$().text().trim(), 'template block text');
});

redpajama_set_name: RedPajamaGithub


October | 2009 | The Daily Grill !!!
Name this Japanese construction company which is one of the world's oldest company operating for over 1,400 years.
Identify this consultancy organization with the image of its founder and its employee portal.
Identify this famous personality from the world of optics.
This media house has its origin in a 1846 agreement between NY City newspapers to share reports from the Mexican-American War. Which?
Under Arm. Strike. Pin. Sports! Identify the company with its logo and this image.
Indian company getting approval for manufacturing & marketing in South Africa. Name?
Name this Thailand-based hotel group which plans to set up 8 hotels in India by 2014.
Identify the Indian Animation Company?
Four Square, Jaisalmer and Red & White. What is the Parent Group called?
What does DB stand in this new telecom company?
What does APM stands here? Which Group does it belong to?
Can't get simple than this! Name the operating system?
Identify the parent company whose logo can be seen in the centre.
Identify this catering company by this ad and punchline.
Connect and identify the Indian partner of this company.
Connect and name the company.
Identify the group with all the clues in the image.
Name the company whose brand has been the biggest Indian Made Foreign Liquor brand.
Connect and name the brand advertised in this "Dil Titli" campaign.
Connect and name the Indian brand associated with this.
Name the orginal company which is now part of Indian Oil.
Identify the company whose MD & CEO can be seen on the bike.
Identify the origin country of these car makers who are about to enter Indian market.
Connect the images and identify the brand.
Who is this entrepreneur, famous in the world of microfinance?
This is a logo of an event which was celebrated in July on a famous personality's birthday. Identify both event & personality.
Identify the media group from its marketing campaign.
This advertising firm is launched by a famous established player in advertising to focus mainly on brand building for entrepreneurial ventures. Who launched it.
He created a tehelka in the world of journalism. Who is he?


Wedding days offer many shooting opportunities and Maryanne + Scott's was no different. The weather during the week leading up to the wedding was wet and at times sunny. It was hard to get a handle on what weather we would experience on the wedding day and in typical Melbourne style the morning was rain soaked whilst the afternoon was filled with the most glorious sun light. I kept thinking to myself how lucky are we?! After many discussions about locations and scouting locations with Maryanne + Scott, Maryanne, had the inspired idea to look at Cruden Farm in Langwarrin and the ground work was set for a wedding full of elegance and beauty.
Cruden Farm is the former residence of the late Dame Elisabeth Murdoch and the grounds were an absolute gift and delight. In fact a photographers dream location. Can I just say, I wanted to spend the whole day at Cruden Farm and keep shooting. It was simply marvellous! The photo's produced with Maryanne + Scott are the result of having a great couple who were easy to work with. As I always say, the more I understand what is important to you about your photo's the better the end result will be, good communication is so important.
So, what else makes for a special day? We already has the palatial grounds, the lake, the stables and who can forget the driveway with the most incredible row of trees I have seen. Well for me the other element that makes a wedding day is the styling and flowers. I feel this was a day where teamwork really came into play. The flowers by Shelly at Melbourne Bridal Rose and Hair & Make-up by Megan at Coastal Hair & Make-up really meant that everything came together beautifully.
Cruden Farm and then the reception at the International In Brighton continued the theme of restrained elegance that was present through out the day. So far this wedding season the bar has been set pretty high. I truly am looking forward to sharing more wonderful weddings like this year.


text:
Interview: Tim Phillips' Inspiring 'Lion Ark' Documents the Most Thrilling Animal Rescue in History
By Danny Miller On December 3, 2013 · Add Comment · In Interviews
Remember the documentary Blackfish from earlier this year that exposed the ugly side of using killer whales for entertainment? After seeing that film, which today was announced as one of the short list of contenders for the upcoming Best Documentary Oscar, I swore off ever returning to places like Sea World. Watching Tim Phillips' great new documentary Lion Ark has produced a similar result: I can never again in good conscience buy a ticket to the Ringling Bros. Circus or any other venture that features wild animals performing tricks for our amusement. You may well feel the same way after you watch this riveting film and see how these animals are treated in such environments. But don't worry, animal lovers, Lion Ark is not just an expose of abuse — with its behind-the-scenes footage of one of the most ambitious animal rescues ever undertaken, I'd go so far as to say it's the feel-good movie of the year!
Director Tim Phillips, along with his wife, Jan Creamer who is seen in the film, are involved with an animals rights group called ADI (Animal Defenders International). The organization spearheaded a shocking undercover investigation in Bolivia that led to a ban on animal circuses in that country. But many circuses defied the law, and kept their captive lions in deplorable conditions as they tried to stay two steps ahead of the authorities. Seeming more like an action-packed thriller than a documentary, Lion Ark documents the daring rescue of every single lion that was still being mistreated in Bolivia as well as the awe-inspiring airlift of these animals to a sanctuary in the United States. It is a story of bravery, compassion, camaraderie, and determination, and the rare animal rights film that will leave you cheering and feeling hopeful. I recently spoke with director Tim Phillips about this important film.
Danny Miller: What an inspiring film, I was on the edge of my seat throughout! The first thing I want to know is how the lions from the film are doing.
Tim Phillips: They're doing fantastically. They've grown a lot, and are so much stronger. The tiny cubs that we rescued are now huge, they've outgrown their parents. When you go to the sanctuary in Colorado, you can see how they've blossomed. They're living in prides, in these big family groups, and as close as possible to the way they'd be living in the wild.
I was stunned to see how quickly the lions improved as soon as you got them away from those appalling conditions in the circuses. Were you shocked at how fast they started adapting to a more human life?
I wouldn't say shocked, exactly, but you can see that even when we made the smallest improvements in their situation, they became so much more engaged with their environment. They became faster, quicker, and more playful, even in the compound in Bolivia, and they began to communicate with each other a lot more. It was as if they were already becoming a pride. It was such a thrill to see them transforming from these lethargic beaten-down creatures into the magnificent animals that they are.
The airlift seemed enormously complicated and risky. Were you worried it wouldn't go as well as it did?
There is always a fear when you're airlifting such animals because you have to cage them up for long periods. Some of the lions had been really starved and their bodies were very damaged. It was a major concern when we were moving them that if one became ill during the ten-hour flight, they could all go downhill very quickly. But happily, they all thrived.
I can only imagine how costly the rescue and airlift were. Was all of that done with the funds you raise in the organization?
Yes, everything we do is funded by public donations. That's what makes it possible for us to put people undercover and expose the findings and secure laws like the ones that were passed in Bolivia. And, of course, these huge rescues. In the film we thank Bob Barker because he gave a very large donation to us that we specifically earmarked for the airlift. Needless to say, clearing an entire country of circus animals is a very expensive thing to do!
The sanctuary in Colorado looks amazing. Is it just not possible to put such animals back into the actual wild?
No, not really. There are lots of reasons that would make that very difficult. When wild animals spend a lot of time living in other countries, you can never really tell what diseases they pick up. They might pick up things that don't necessarily affect lions but can massively damage other parts of an eco-system if they were sent back to their original countries. And these animals have never hunted, it's very doubtful that they'd be able to survive on their own.
The actual collection of the animals from various circuses is such a suspenseful part of the film. Did you guys ever worry for your safety in those situations? Some of these people seemed like rather unsavory characters!
There were a few tense moments and several times when the authorities warned us that certain circuses said they were going to attack me and Jan and beat us up. It didn't come to that but we occasionally brought police with us. There was often a lot of shouting and shoving and pushing about but generally we got through those situations and things calmed down.
You obviously had the law on your side. Was it the threat of facing jail or huge fines that eventually made them relent?
Yes, I think that was ultimately what won through. These people were definitely defying the law. There was this one circus that kept constantly moving to avoid us. We placed one of our people there undercover working as a clown so from then on we always knew where they were! We just moved into the town one night and raided them early in the morning. But yes, sometimes it got very tense.
After seeing the film Blackfish I decided I could never again go to Sea World. This film made me realize that I can no longer support circuses that feature animals even though I grew up enjoying those and used to frequently take my kids to the circus. Do you think there is any justification for having animals be involved in such activities or do you think all such circuses need to come to an end?
I think they have to come to an end. With the circuses, in particular, no other animals have to spend almost their entire lives in such temporary accommodations. That will always be a huge problem. And then having to do all these shows and tricks which leads to constant contact with people and physical violence because they are dangerous animals and so these people have to dominate them because they're working with them in such restricted space. When you stop and think that that's just in the name of a few minutes of entertainment, it just has to stop.
Have you been able to help pass laws banning animal circuses in other countries?
Oh, yes, there's been an unprecedented wave of countries that are now banning the use of wild animals in circuses. In Asia you've got countries such as China and Singapore and Taiwan. In Europe, you have Portugal, Greece, the Czech Republic, and now Great Britain has some legislation drafted. In South America, there is Bolivia, Paraguay, Peru, Ecuador, and Columbia, and in Central America, El Salvador and Costa Rica.
Wow, that's amazing.
The 25 countries who have passed such laws are all very different socio-economically and yet they're all saying, "No this shouldn't happen, we should not be treating animals like this just for entertainment."
Where do things stand in the United States?
There was some legislation presented this year that didn't pass. We're hoping for more support next year. It's unlikely it will get through on its second go, but it will happen eventually. There is no longer any real appetite in the U.S. for animals in circuses, the few that still have them have declining audiences and there are now many successful circuses without animals. I think there's a lot of passion for this issue in the U.S. It's no accident that the lions from Bolivia found refuge in sanctuaries in this country.
But you must have received push-back from places like Ringling Brothers and other big corporations.
Yes, we do get that and we have engaged in debate with those people in the press and in town hall meetings. But the amount of violence that has been caught on film, including at the big-top circuses, is just staggering. So much of that violence shows a culture of intimidation and that these animals are forced to live in terribly limited environments. That's why it's got to change.
I'm sure there are many Americans who want to think, "Well, maybe that goes on in these other countries, but surely not here in the United States!"
Yeah, we've done a lot of Q&As at screenings and film festivals and people always ask what the difference is between what they see in the film and what's going on in circuses here. Often it's just that the cages look shinier — they're clean, they're freshly scrubbed, but in the end do the animals actually have better lives? Not really. ADI did an exposé on a group of bears that are currently touring the United States. We followed them and found them living most of the time in this awful truck. They're taken out to do their show — ride bikes, do things that are really archaic, it seemed like something from 150 years ago — and then put back into the truck after the show. It was like they were living in dungeons, these little tiny pens, even if they were shiny, clean, and sanitized. I would say that life for those bears was even worse than for the lions in Bolivia, even the eight you see in the film that were living in that small disgusting cage — at least they were able to look out into the world. The bears in America have absolutely nothing.
Are you also against zoos, or does it depend on how they're set up?
There's obviously a huge variation out there. We certainly think that deliberately taking any animals from the wild or deliberately breeding them for incarceration is wrong. Doing that simply to give people a chance to see these animals is not a good reason to take everything of value from their lives. But the truth is that we're always going to be stuck with wild animals in captivity to some extent. Animals taken from drug dealers or rescued countries like Bolivia, and so forth, but there needs to be a real shift towards keeping these animals in a way that suits them. We need to have more sanctuaries that specialize in certain species so that we can do the best for those species. I think the Lion Ark lions highlight how you can still have animals that you can go and visit but it's not intrusive, they've got lots of space, and they live very happily in appropriate family groups.
Click here for more information on Lion Ark including upcoming screenings.
Tagged with → ADI • Animal Rights • Documentary • Jan Creamer • Lion Ark • Tim Phillips
'American Hustle' Dominates New York Film Critics Circle Videophiled: Boxes of Movies for the Holidays – 'Argo,' 'Anchorman,' Jack Ryan and more

redpajama_set_name: RedPajamaCommonCrawl


It has taken a while for South Florida to catch up with the worldwide green building LEED (Leadership in Energy and Environmental Design) certified trend. Since its inception in 1998, the U.S. Green Building Council has strived to move construction technology forward and improve efficiency in buildings for a more sustainable and healthy environment. Despite this, Florida didn't get its first LEED certified building until 2009.
However, in the last three years LEED certified green buildings have begun to emerge, with 60+ buildings achieving LEED certification in 2012. Among some of the notables are 1450 Brickell tower, 600 Brickell tower – which was the first building to achieve the Platinum LEED certification in Florida, – and the Florida Institute for Neuroscience, among others. The recent real estate crash could be to blame for the slow adoption of the new technologies, but many other factors come into play as well. To achieve LEED certification, buildings have to implement technologies that are not commonly used and make construction more expensive, but companies have realized that the long term benefits outweigh the initial cost (Greg Kats). Furthermore, as LEED building technologies become more widely available, it is expected that implementation can become a less costly endeavour and become an industry standard across the board.
These are the baby steps that Florida and many other states and countries are taking to make for a more sustainable and healthy world, as the USGBC strives for the Architecture 2030 challenge, which aims to reduce building's fossil fuel energy consumption to 0% by 2030.
For more information on the Architecture 2030 challenge click here.


text:
New Pre-Packaged Resolution Process Not Beneficial for Micro, Small MSMEs: Claim Experts
Prudhviraj Rupavath | 13 Apr 2021
The majority of the unregistered MSMEs to the tune of six crore are outside the ambit of the IBC Code.
Representational use only. Image Courtesy: Livemint
The pre-packaged insolvency resolution process for Micro, Small and Medium Enterprises (MSMEs) specified under the new Insolvency and Bankruptcy Code (Amendment) Ordinance, 2021, doesn't address the real problems of micro and small enterprises, say experts. Further, the majority of the MSMEs don't fall inside the ambit of the new Code as per several provisions in the Ordinance.
On April 4, President Ram Nath Kovind promulgated the ordinance to make amendments in the Insolvency and Bankruptcy Code, 2016 (Code), laying out procedures for the resolution of bankrupt MSMEs.
It is expected that the incorporation of the pre-packaged insolvency resolution process for MSMEs in the Code, states a press statement by the Ministry of Corporate Affairs, will alleviate the distress faced by MSMEs due to the impact of the pandemic and the unique nature of their business, duly recognising their importance in the economy.
"It provides an efficient alternative insolvency resolution framework for corporate persons classified as MSMEs for timely, efficient and cost-effective resolution of distress thereby ensuring positive signal to debt market, employment preservation, ease of doing business and preservation of enterprise capital," according to the Ministry.
This ordinance is simply a glorified one time settlement scheme for MSMEs although through a cumbersome process especially for micro and small enterprises, said K E Raghunathan, convenor of the Consortium of Indian Associations.
The central government has set the threshold default of Rs 10 lakh for initiating the resolution process for MSMEs and the time frame for the process is fixed at 125 days.
"The resolution process in the National Company Law Tribunal (NCLT) comes with a cost, making it difficult for the micro and small scale entrepreneurs. The package does not provide an opportunity for redressal mechanisms for bankrupt MSMEs," said Raghunathan.
As per the ordinance, classification of MSMEs under sub-section (1) of section 7 of the Micro, Small and Medium Enterprises Development Act, 2006, is required for initiating the pre-packaged resolution process. It is to be noted that only 26.42 lakh have been registered under Udyam registration (new registration process specified by the central government). Reportedly, another 76 lakh MSMEs which have earlier registered with the government did not re-register under Udyam by March 31, a cut off date specified by the central government. Industry bodies are appealing to the central government to extend the deadline and advertise the benefits of the registration among industries.
While there is no recent government data on the number of MSMEs, according to the National Sample Survey (2015-2016), there are over 6.3 crore unincorporated MSMEs in India. As the numbers specify, the majority of the unregistered MSMEs to the tune of six crore are outside the ambit of the IBC Code.
Furthermore, the resolution package is restricted to only companies and Limited Liability Partnerships while entities registered as sole proprietorship, partnerships and Hindu Undivided Family forms of MSMEs are excluded from its ambit.
"Extending a bankruptcy resolution scheme under the IBC to unincorporated entities and proprietorships, however, is hugely challenging given the large number of such enterprises and the limited infrastructure including the number of National Company Law Tribunal (NCLT) benches," the Mint quoted a government official on the condition of anonymity.
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Hwy. 185 crash claims life
BY BRUCE BRANUM
On Monday, March 30, at approximately 6:25 p.m. Capricia LeeAnn Jackson, age 37, of Butler County was killed in a car wreck on Alabama State Highway 185, approximately five miles south of Greenville.
According to a press release by Sgt. Drew Brooks, who is with ALEA and Troy Post Commander for the Highway Patrol / Troop B, Jackson died from injuries sustained during the crash which only involved one vehicle.
Jackson was driving a 2007 Lincoln Town Car and if left the roadway and struck a tree, according to Sgt. Brooks.
Sgt. Brooks stated, "Ms. Jackson was transported to a local hospital where she later died from the injuries sustained during the crash. Speed is believed to be a factor in the crash. Nothing further is available as Alabama State Troopers continue to investigate."


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Anthony Bellanger, né le  au Mans (France), est un journaliste, historien (docteur en histoire) et syndicaliste français.

Depuis , il est le secrétaire général de la Fédération internationale des journalistes (FIJ). Depuis 2020, il est également professeur invité de journalisme à l'école de journalisme de l'université catholique de Louvain.

Membre fondateur de l'association basée à Bruxelles En-GAJE, ENsemble Groupe d'Aide aux Journalistes Exilés, il a participé à la création du média en ligne, LATITUDES, dont il fut le premier rédacteur en chef en 2022.

Biographie

Formation
Après un baccalauréat littéraire au lycée Montesquieu du Mans, Anthony Bellanger entre à l'université du Maine et obtient, en 1995, une licence d'histoire. Il poursuit son parcours à l'université d'Angers où il obtient une maîtrise, en 1996, puis un diplôme d'études approfondies (DEA) deux ans plus tard, avant d'entamer une thèse de doctorat en histoire du Moyen Âge qu'il soutient en  à Angers sous la direction des professeurs Noël-Yves Tonnerre et Michel Pastoureau.

Anthony Bellanger est docteur en histoire médiévale. Sa thèse a été publiée en deux tomes par les Éditions universitaires européennes.

Carrière professionnelle
Entre 1996 et 1998, Anthony Bellanger effectue son service national comme objecteur de conscience en tant que chercheur aux musées d'Angers. Après son inscription en thèse de doctorat en 1998 et jusqu'en 2001, il se voit proposer un poste de chargé de cours d'histoire médiévale à l'université d'Angers.

Parallèlement, en 1998, il intègre la rédaction du quotidien régional Le Courrier de l'Ouest, titre du groupe Ouest France, qu'il quittera en 2014 pour rejoindre la FIJ à Bruxelles.

En 2004 et 2005, il fonde puis dirige la publication du magazine trimestriel national français L'Œil du patrimoine.

Depuis 2014, tout en assumant ses missions de secrétaire général de la FIJ, Anthony Bellanger collabore en tant que journaliste indépendant pour divers titres de la presse internationale, notamment Le Monde Diplomatique, Mediapart et L'Humanité.

Anthony Bellanger partage son expérience internationale régulièrement auprès des jeunes scolaires ou des étudiants dans l'enseignement supérieur, notamment à l'UC Louvain (UCL), l'Université libre de Bruxelles (ULB), l'École supérieure de journalisme de Paris, l'université catholique d'Angers...

Depuis 2020, il est professeur invité de journalisme à l'école de journalisme (COMU) de l'UC Louvain.

En 2021, membre du comité éditorial du média en ligne LATITUDES, Anthony Bellanger a été nommé rédacteur en chef en 2022 lors du lancement du site en ligne consacré aux thèmes des droits humains, des migrations et de la liberté d'expression. Ses contenus sont réalisés exclusivement par des journalistes en exil, membres de l'association En-GAJE, des étudiants en journalisme de l'ULB et de la VUB, les deux universités bruxelloises francophone et néerlandophone.

Activités syndicales et internationales
Adhérent du Syndicat national des journalistes (SNJ), il exerce depuis 2003 différents mandats en entreprise (délégué syndical, délégué du personnel, élu au Comité d'entreprise et au CHSCT). Il est élu au Comité national du SNJ en 2005, au Bureau national en 2008, avant de devenir en 2010 secrétaire général, puis, en 2011, Premier secrétaire général au congrès d'Autrans. Il est réélu à cette fonction en 2013 au congrès d'Angers avant de démissionner le , lors du bureau national pour prendre la direction de la Fédération internationale des journalistes (FIJ) à Bruxelles.

Comme Premier secrétaire général du SNJ, il s'est notamment exprimé sur l'autorégulation déontologique des journalistes ou sur la situation générale de la profession. Lors de la polémique sur le « mur des cons », il apporte le soutien du SNJ au Syndicat de la magistrature.

Le , il rejoint la Fédération internationale des journalistes (FIJ), qui siège à Bruxelles, dont il devient Secrétaire général adjoint, puis Secrétaire général le . Il devient ainsi le deuxième Français depuis 1926 à occuper cette fonction. La FIJ, créée à Paris en 1926, a son siège social à Bruxelles depuis 1952 et a ouvert des bureaux régionaux à Buenos Aires (Argentine) pour l'Amérique latine et les Caraïbes, à Dakar (Sénégal) pour l'Afrique et à Sydney (Australie) pour l'Asie-Pacifique. La FIJ, première organisation mondiale de la profession, représente en 2023 plus de  répartis dans  du monde.

En tant que secrétaire général de la FIJ, Anthony Bellanger est aussi le porte-parole et le défenseur de l'ensemble de la profession dans le monde, notamment à l'Organisation internationale du travail (OIT) à Genève, à l'Unesco à Paris ou à l'Organisation des Nations Unies (ONU) à New York. Lors du  mondial de la FIJ à Tunis (11-), une nouvelle Charte mondiale d'éthique des journalistes a été adoptée à l'unanimité par les délégués; des travaux qu'il a coordonnés avec les élus et experts de la FIJ entre 2018 et 2019.

Références

Liens externes

Syndicaliste français
Journaliste français du XXIe siècle
Naissance au Mans
Naissance en août 1973
Objecteur de conscience français

redpajama_set_name: RedPajamaWikipedia


Overseas network October 13th According to the Japanese Asahi Shimbun, a French Airways Flight 293 (Boeing 777) from Japan's Haneda Airport to Paris, France, late at night, deviated from the usual route after take-off, and flew through downtown Tokyo at low altitude. The area even flew over the Japanese Imperial Palace.
In general, aircraft taking off from Haneda Airport, in order to avoid noise pollution in the center of Tokyo, are required to fly back to Tokyo Bay shortly after take-off. Japan's Ministry of Land, Infrastructure, Transport and Tourism called the incident a rare case and asked the airline to explain it.
The Japanese government's regulations specifically include: the plane takes off from Haneda Airport, and if it needs to go north, it must swing rightward within about 7.4 kilometers. Japan's Ministry of Land, Infrastructure, Transport and Tourism said that the flight was about 11:13 pm and took off at the Haneda Airport C runway. It went straight north and passed through Shinagawa Station at 11:14.
The Air France flight over the Japanese Imperial Palace at 11:15, and flew out of Tokyo Bay at 11:16 after passing over the Arakawa River in Edogawa. It is reported that the Ministry of Land, Infrastructure, Transport and Tourism has not received any complaints about noise.


This is a pretty loaded question, but honestly, it's one of my favorite things to help clients with. I'll step you through how to evaluate the time and timing constraints and balance that with varied interests of the people you want to travel with.
1. How much TIME do you have to travel?
Regardless of the number of days you have available to travel, you want to make the best use of that time as possible.
The further the destination, the less likely you will be to return to that region as frequently as other places. Therefore, my rule of thumb is the further away a place is, the longer you should spend there to get the most value of your time there.
I like to plan out my big trips first (i.e. 1 week or longer) and then fill in the gaps with weekend getaways. Remember, you aren't constrained to going to just one place – use your vacation days creatively to maximize your time off!
Most places in the world have high season, shoulder season, and low season. As much as I love Chicago, if I'm talking to a person that lives in southern California I would NEVER recommend they come to Chicago in January! Clearly, summer is the high season in Chicago, and for good reason. However, there are a few places, like Hawaii, where it's ALWAYS high season (with Christmastime being the highest of the high season!).
A quick google search for "best time to visit" + the destination you are considering will give you results to help you figure out when high, shoulder, and low travel seasons are there.
Generally speaking, you get better deals during low and shoulder season and you aren't battling the large crowds of high season, however you may not get the best weather. I highly recommend traveling in shoulder season when you can since it's a bit of a balance of both worlds.
Also, remember that destinations in the Southern Hemisphere have opposite seasons, so right now places like Australia, New Zealand, Argentina, and South Africa are enjoying the height of their summer. This makes me want to hop on a plane to one of these places immediately!
If you can only travel during certain times of the year based on work or school schedules, you may have to go to some places during high season and that is ok as long as you are prepared for that from a budget and expectations standpoint.
My biggest piece of advice for those that have to travel during high season is to BOOK EARLY! Like a year early, no joke. If you are thinking about a summer vacation in a popular place like Europe, you need to get on it! The best places to stay at the lowest rates are going quickly, so all is not lost, but I highly recommend taking action quickly in order to ensure the best overall travel experience.
3. Who are you going with and what do each of you want to EXPERIENCE?
Who are you planning on going on this trip with? If it's a solo trip – awesome, you get to call the shots on the types of experiences you want to have! However, if you're traveling with a significant other, your kid(s), or a group, then there are others to consider.
Everyone travels differently, so the trick is to find a well-balanced destination where everyone has a little something they can get excited about doing, seeing, eating, and/or learning. Trust me, if the people you are with are not enjoying their vacation, then YOU will not enjoy your vacation!
Take a cooking class and eat ALL the food?
Lounge on a beach with a drink in hand?
Volunteer with an organization you're passionate about?
Do something physically challenging, like summit a mountain?
Get their adrenaline running with an adventurous escape?
Learn about history in the place it happened?
Experience zen in a yoga retreat?
The possibilities are endless, but it's important to discuss this up front so you're on the same page.
Take a look at your list of possible destinations – are there any that pop out at you that check the interest boxes of everyone you're going on the trip with?
If you've narrowed down your list, check out how to figure out what to budget for your trip when comparing different destinations.
Comment below where you're going this year!


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Please show me example sentences with callow.
It is a synonym of inexperienced. However, I have never seen anyone write or speak this word before, so I wouldn't say that there are any common phrases with it.
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Callow


or the neighbors on your block. But whoever YOUR people are, summer is for face-timing them the old fashioned way...in person!
But the question is always: "Where are we going to meet?"
A few weeks ago Edith, a Bureau co-worker, attended a family party for 30 people on the outdoor deck of Stingers. She gave a big thumbs up to the event, the service, great river views and patio breezes at Kevin Farrell's and Chef/Owner Joseph Mele's newest venture at the Ridley Marina.
Bring friends for lunch, dinner, or just enjoy some cocktails, as you watch the boats and local wildlife from the nearby John Heinz Refuge at this beautiful waterfront establishment.
AUTOGRAPH BRASSERIE IS HOME TO GREAT FOOD... and exudes a sublime legacy of celebrity!
Marilyn, another team member here at the Bureau, joined a group of old friends last month at Autograph Brasserie, on the Main Line in Wayne. She gave it rave reviews for atmosphere, food and decor! Autograph Brasserie, part of Marty Grims' Fearless Restaurants, opened in 2016 in the former location of Peppercorn and George's. The building's rich history and French influence framed the inspiration of a concept Grims has had for over 20 years.
Featured throughout the restaurant are over 1000 celebrity autographs, music albums, and authentic photographs celebrating those individuals that have impacted popular American culture.
THE PATIO: Enjoy outdoor dining on their hidden patio featuring a bubbling fountain and lush florals.
HAPPY HOUR: Monday – Friday, 4 to 7 PM with $5 specially priced cocktails, wine & beer and Happy Hour menu featuring $5 snacks, $7 small plates and $10 medium plates.
WINE NOT SUNDAYS: Half priced bottles every Sunday for Brunch and Dinner.
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FREE FUN FOR GROUPS LARGE & SMALL: Rose Tree Park's Summer Festival is the biggest free party in the County, and it comes with stars overhead and great entertainment on the stage.
Just two weeks left to enjoy the best entertainment deal in town! Come early and toss a ball around, or just share blankets and catch up with everybody as the band sets up to play.
Aug 11 Blackbird Society Orchestra, the roaring 20's Roar Again!
All Performances begin at 7:30pm -- And please bring blankets and chairs for seating. Show will be cancelled in the event of rain.
HOW ABOUT A DAY AT THE RACES? Meet at Harrah's Philadelphia Casino & Race Track!
Harrah's offers full-card Philadelphia simulcasting of both thoroughbred and standardbred signals, seven days a week, and the Casino provides a state-of-the-art indoor wagering area with plenty of seating and simulcast video options.
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In earliest days, there were reading of Richard Wurmbrandt and far right leanings especially with regard to patriarchical control with women to be submissive to men. But truly, from the core, people- men and women were all to be submissive to Berg who was presented as someone that God talked to in visions and dreams. When witnessing to youth of the day who were gathered from the streets and universities, it was popular in the day to admire leftist policies amongst the youth. Berg wanted young followers and so he used Acts 2:44 and 45 as Marxism something that Jesus came up with. In reality, it was a perfect recipe for control because you "forsake all" and give up all your money and property and yourself, and then it is redistributed by leadership. Some people came in with little to nothing but the second line of income was through parents and relatives who were sent highly sanitized missionary newsletters and writings which were censored and people were re-directed regarding what and how to write, what not to write and how to ask for money from home. Unless you were a top leader, whatever was sent did not go to you. It went to leadership unless it was something for which there was no need. Then you might be able to keep it. Whatever leadership took, you were instructed to write home and tell them the package was received and the cookies (that leadership ate) or the funds or whatever was sent was received by you. If any literature came in or something the leadership felt was inappropriate, it was taken and discarded. His handful of charismatic leaders parroted what Berg taught, that God's form of government was a DICTATORSHIP. And that the recipe was to obey without questioning, whatever leadership told us. There were tests for this. It was a very exploitative society and hardly a left wing society. Berg also had lots of end time paranoia so there were skits and sessions to act out what that time might be like. Lots of control by fear. Birth control was not allowed. Dating was not allowed and a couple couldn't start thinking about getting together until they were put together. It was a parasitical dictatorship more like Idi Amin.
Berg was initially very pro-Israeli until he went to Israel and was rejected finding out he could not start his dream Kibbutz there. He was very much in the vein of the Jews being the chosen people. Rejection of him and his message meant they became demonized. Then he became pro Arab and Ghaddafi.
There was a point in time when South Africa was still practicing Apartheid and whites were solidly in control, that he thought that may be a place to go when things in the world started heating up. Certainly, that was not a leftist society.
He idealized Europe for a time, but once his insane doctrines started wreaking havoc world wide press got pretty negative. He moved towards third world countries..and on and on.
He as a chameleon and a megalomaniac, so he fed off of whatever host accepted him, then demonized them when he (or his followers) was ousted. I think he may have gone left when he felt his dreams and visions indicated that USA and western world would be destroyed and the rulers of the future would be Russia. He felt a need to seek out those hosts. He also sought out crazy azz Ghaddafi who he thought was likely the False Prophet. Not exactly a left winger, but a despot. He was accepted there and then rejected.
Early family times, women couldn't wear make up or shave. Later family he turned women out as prostitutes. He was all about contradictions, but all along, an opportunist parasite.