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99 lines
3.0 KiB
TeX
99 lines
3.0 KiB
TeX
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8x]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage{pgfplots}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usetikzlibrary{automata,matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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\usepackage{multirow}
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\usepackage[german]{fancyref}
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\polyset{style=C, div=:,vars=x}
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\pgfplotsset{compat=1.8}
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\pagenumbering{arabic}
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% ensures that paragraphs are separated by empty lines
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\parskip 12pt plus 1pt minus 1pt
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\parindent 0pt
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% define how the sections are rendered
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\def\thesection{4.\arabic{section})}
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\def\thesubsection{\arabic{subsection}.}
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\def\thesubsubsection{(\roman{subsubsection})}
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% some matrix magic
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\begin{document}
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\author{Benjamin Kuffel, Jim Martens\\Gruppe 6}
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\title{Hausaufgaben zum 10. November}
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\maketitle
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\setcounter{section}{2}
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\section{} %4.3
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\subsection{}
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\begin{alignat*}{2}
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L(TS_{kuchen\_teil}) &=& (v(htwb^{*}k)^{*}o + r)^{*} \\
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L^{\omega}(TS_{kuchen\_teil}) &=& (v(htwb^{*}k)^{*}o + r)^{\omega}
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\end{alignat*}
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\subsection{}
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\[
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SS(M) = 1(3(5763)^{*}1 + 1)^{\omega}
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\]
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\subsection{}
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\begin{alignat*}{2}
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E_{S}(M) &=& E_{S}(1(3(5763)^{*}1 + 1)^{\omega}) \\
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&=& (E_{S}(1)(E_{S}(3)(E_{S}(5)E_{S}(7)E_{S}(6)E_{S}(3))))^{\omega}
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\end{alignat*}
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\subsection{} %
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\begin{alignat*}{2}
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Sat(Teig \vee \lnot Hitze) &=& \{1, 4, 5\} \\
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Sat(\lnot Teig) &=& \{1, 2, 3, 6, 7\} \\
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\intertext{Auflösen der Implikation}
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f &=& GF((Teig \vee \lnot Hitze) \Rightarrow F \lnot Teig) \\
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&\Leftrightarrow & GF(F(\lnot(Teig \vee \lnot Hitze) \vee F \lnot Teig))
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\intertext{Berechnung der weiterführenden \(Sat\)-Mengen}
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Sat(\lnot(Teig \vee \lnot Hitze)) &=& \{2,3,6,7\} \\
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Sat(\lnot(Teig \vee \lnot Hitze) \vee \lnot Teig) &=& \{1,2,3,6,7\}
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\end{alignat*}
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Da der Anfangszustand in der Menge vorhanden ist, gilt die Formel.
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\subsection{} %Gegenbeispiel finden
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\begin{alignat*}{2}
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Sat(Backen) &=& \{6,7\} \\
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Sat(Zeit) &=& \{2,6\} \\
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Sat(Backen \wedge Zeit) &=& \{6\}
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\end{alignat*}
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Die Formel gilt nicht im Anfangszustand. Gegenbeispiel:
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\[
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\pi = 14576(32)^{\omega}
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\]
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Pfad, bei dem es funktioniert:
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\[
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\pi = 1457(6)^{\omega}
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\]
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\section{} %4.4
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\begin{tabular}{l|l|l}
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\(f\) & \(M_{kuchen} \models f \) & \(M_{kuchen}, \pi \models f\)\\
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\hline
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\(\lozenge \square(\lnot Teig \vee Hitze) \)& nein & ja \\
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\(\square \lozenge(\lnot Teig \vee Hitze)\) & ja & ja \\
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\(\square (Hitze \;U\; Backen)\) & nein & nein \\
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\(\square \lozenge (Backen \Rightarrow XX\lnot Backen)\) & ja & ja \\
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\(\square ((Hitze \wedge Teig) \Rightarrow \lozenge \lnot Teig)\) & ja & ja \\
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\(XX \lozenge (\lnot Hitze \wedge \lnot Teig)\) & nein & nein
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\end{tabular}
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\end{document}
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