112 lines
3.9 KiB
TeX
112 lines
3.9 KiB
TeX
\documentclass{article}
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\usepackage{xspace}
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\usepackage{hyperref}
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\usepackage{bm}
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\usepackage[english]{babel}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{mathtools}
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\newcommand{\jh}[1]{{\leavevmode\color{blue!50!red}#1}}
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\input{notation.tex}
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\input{glossary.tex}
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\pagenumbering{arabic}
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\pagestyle{plain}
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\newtheorem{theorem}{Theorem}
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\newtheorem{corollary}{Corollary}
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\newtheorem{lemma}{Lemma}
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\newtheorem{proposition}{Proposition}
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\begin{document}
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% \maketitle
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\definition{monotone}{define monotone}
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\definition{image}{define set image}
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\definition{lubglb}{define glb and lub}
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\definition{topbot}{define $\top$ and $\bot$}
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\definition{completelattice}{define complete lattice}
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Hello world\cite{tarskilatticetheoretical1955}
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First, we generalize Knaster-Tarski Fixpoint Theorem.
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$$\fixpointsOf(S)$$
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\begin{theorem}[Tarski-Knaster Fixpoint Theorem~\cite{tarskilatticetheoretical1955}]\label{tarskitheorem}
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For a \Monotone function $o$ over a \CompleteLattice $\langle \L, \lte \rangle$, we have that $\langle \fixpointsOf(o), \lte \rangle$ is a \CompleteLattice.
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\end{theorem}
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\begin{theorem}
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Trivial attempt using set preimage / image to attain LUB / GLB in image lattice does not work.
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Consider antichain of 2 that map to a chain in image.
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The antichain's LUB does maps to a third, distinct element which extends the chain to three in the image
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If we attain LUB of preimage, then its not the least element :)
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\end{theorem}
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\begin{theorem}\label{imagelattice}
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For a \Monotone function $o$ over a \CompleteLattice $\langle \L, \lte \rangle$, we have that $\langle o\imageNoLink{\L}, \lte \rangle$ is a \CompleteLattice.
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\end{theorem}
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\begin{proof}
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Consider $\langle \L', \lte \rangle$ where
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\begin{align*}
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\L' \define \{ x' ~|~ x \in \L \} \\
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x' \lte y' \iff x \lte y \textrm{ where } x, y \in \L
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\end{align*}
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Clearly, $\langle \L', \lte' \rangle$ is a \CompleteLattice.
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Combining $\L$ and $\L'$, we formulate $\L^*$:
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\begin{align*}
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\L^* &\define \{ \top^*, \bot^* \} \union \L \union \L' \\
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\forall x^*, y^* \in \L^*, x^* \lte y^* &\iff \begin{cases}
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(x^* = \bot^*) \lor (y^* = \top^*), \\
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\textrm{or } x, y \in \L, \\
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\textrm{or } x', y' \in \L'
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\end{cases}
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\end{align*}
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It is easy to show that $\langle \L^*, \lte \rangle$ is a \CompleteLattice.
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Consider the operator $o^*$ that maps elements from $\L$ to $\L'$ and every element in $\L'$ to itself.
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\begin{align*}
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o^*(x) \define \begin{cases}
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o(x)' & \textrm{if } x \in \L \\
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x & \textrm{otherwise}
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\end{cases}
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\end{align*}
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We have $o^*\image{\L \union \L'} \subseteq L'$ and $\fixpointsOf(o^*) = o\image{\L}' \union \{ \bot^*, \top^* \}$.
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By Tarski-Knaster Fixpoint Theorem (Theorem \ref{tarskitheorem}), $\fixpointsOf(o^*)$ is a \CompleteLattice.
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Thus, $o\image{\L} \union \{ \bot^*, \top^* \}$ is a \CompleteLattice. For any $S \subseteq \L$, we have $\glb S \not\in \{ \bot^*, \top^* \}$ and $\lub S \not\in \{ \bot^*, \top^* \}$, thus $\langle o\image{\L}, \preceq \rangle$ is a \CompleteLattice.
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\end{proof}
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The following follows directly from the contrapositive of Theorem \ref{imagelattice}.
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\begin{corollary}
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If $\langle o\image{\L}, \lte \rangle$ is not a \CompleteLattice, then $o$ is not \Monotone.
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\end{corollary}
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\begin{definition}
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An approximator set $H$ captures a nondeterministic approximator $o$ if for each consistent pair $(x, y)$
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\begin{align*}
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(H\image{(x, y)}_1, H\image{(x, y)}_2) = o(x, y)
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\end{align*}
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\end{definition}
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\begin{theorem}
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A nondeterministic approximator $o$ has a set of
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\end{theorem}
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\begin{theorem}
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For any nondeterministic approximator $o$, there exists an approximator set $H$ s.t.\ gamma stable models correspond to n-stable models
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\end{theorem}
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\begin{proof}
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foo
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\end{proof}
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\bibliographystyle{plain}
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\bibliography{references}
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\end{document}
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