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finalized Pumping Lemma
changed titlelayout and pagebreaks in multiple documents
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@ -6,7 +6,7 @@
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\title{Ko-Rekursion/-Induktion}
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\title{Ko-Rekursion/-Induktion}
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\date{ }
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\date{ }
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\begin{document}
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\begin{document}
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\maketitle
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%\maketitle
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\section{Korekursion}
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\section{Korekursion}
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\textbf{Gegeben:}
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\textbf{Gegeben:}
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\begin{fleqn}
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\begin{fleqn}
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@ -4,10 +4,11 @@
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\usepackage{ulem}
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\usepackage{ulem}
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\DeclareMathSizes{10}{10}{10}{10}
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\DeclareMathSizes{10}{10}{10}{10}
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\setlength{\parindent}{0pt}
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\setlength{\parindent}{0pt}
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\title{Ko-Rekursion/-Induktion}
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\title{Polynomordnung}
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\date{Oktober 2015}
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\date{}
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\begin{document}
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\begin{document}
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\maketitle
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%\maketitle
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\section*{Polynomordnung}
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\section{In Funktionsschreibweise bringen}
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\section{In Funktionsschreibweise bringen}
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Infixnotation:
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Infixnotation:
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\[x \uparrow ( y \uparrow z) \rightarrow_{0}\; x \uparrow (y \downarrow y)\]\\
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\[x \uparrow ( y \uparrow z) \rightarrow_{0}\; x \uparrow (y \downarrow y)\]\\
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@ -2,16 +2,19 @@
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\usepackage{amsmath}
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\usepackage{amsmath}
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\usepackage{ amssymb } % >= as \geq
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\usepackage{ amssymb } % >= as \geq
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\usepackage{nccmath}
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\usepackage{nccmath}
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\usepackage{upgreek}
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%New Commands
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%New Commands
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\newcommand{\xhspace}[0]{\noindent\hspace*{5mm}}
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\newcommand{\xhspace}[0]{\noindent\hspace*{5mm}}
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\DeclareMathSizes{10}{10}{10}{10}
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\DeclareMathSizes{10}{10}{10}{10}
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\setlength{\parindent}{0pt}
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\setlength{\parindent}{0pt}
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\title{Pumping Lemme f\"ur Regul\"are Sprachen}
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\title{Pumping Lemma f\"ur Regul\"are Sprachen}
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\date{ }
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\date{ }
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\begin{document}
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\begin{document}
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\maketitle
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\section*{Pumping Lemma f\"ur Regul\"are Sprachen}
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%\maketitle
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%----------------------------------%
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%----------------------------------%
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\textbf{L ist regul\"ar wenn:}
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Konkret geht es in diesem Beispiel um eine Sprache die alle ge\"offneten Klammern auch wieder schlie$\upbeta$en soll, oder allgemeiner, um eine Sprache die sich eine Anzahl merken muss.\\\\
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\textbf{'L' ist regul\"ar wenn:}
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\[
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\[
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\forall l \geq 1\;.\, \exists w \in L\;mit\; |w|\geq l
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\forall l \geq 1\;.\, \exists w \in L\;mit\; |w|\geq l
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\]
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\]
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@ -22,9 +25,40 @@
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\xhspace $|$v$|$ $\geq$ 1\\
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\xhspace $|$v$|$ $\geq$ 1\\
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\xhspace $|$uv$|$ $\leq$ l\\
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\xhspace $|$uv$|$ $\leq$ l\\
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\textbf{gilt:}\\
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\textbf{gilt:}\\
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\xhspace u$v^{k}$z $\notin$ L
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\xhspace u$v^{k}$z $\notin$ L \\ \\
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\textbf{Beweis, dass 'L' \underline{NICHT} regul\"ar ist}\\\\
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\begin{tiny}
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Sei l $\geq$ 1 gegeben, w\"ahle:
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\copyright\ Joint-Troll-Expert-Group (JTEG) 2015
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\[
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\end{tiny}
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w\;=
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\underbrace{(\,(\,(\,...\,(}_{l+k\;mal}
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\hspace{10mm}
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\;a
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\hspace{1cm}
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\underbrace{,\,b\,)\,,b)\,...\,)}_{l+k\;mal}
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\hspace{10mm}
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\in L
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\]
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k als feste Zahl w\"ahlen, z.B. 10:
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\[
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w\;=
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\underbrace{(\,(\,(\,...\,(}_{l+10\;mal}
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\hspace{10mm}
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\;a
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\hspace{1cm}
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\underbrace{,\,b\,)\,,b)\,...\,)}_{l+10\;mal}
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\hspace{10mm}
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\in L
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\]
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\textbf{dann gilt:}\\\\
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\noindent \hspace*{1cm}$\forall$ uvz \hspace{5mm}\textbf{mit}\hspace{5mm}w = uvz
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\\\\
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\noindent \hspace*{1cm}u = $(^k$ \\ \\
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\noindent \hspace*{1cm}v = $(^m$\\ \\
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\noindent \hspace*{1cm}z = $(^{[l+10-k-m]}$
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\hspace{3mm}a\hspace{3mm},b\,)\,,\,b\,)\,...\,,\,b\,)\\\\
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\textbf{dann w\"ahlenn wir:}\\\\
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\noindent \hspace*{1cm} w' = u$v^0$z = uz $\notin$ L\\\\
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Denn es gehen [k+l+10-k-m = 10+l+m] Klammern auf und nur [l+10] Klammern zu.
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Da m$\geq$ ist [10+l+m $>$ 10+l].
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D.h. w' ist nicht in 'L' und 'L' damit nicht regul\"ar.
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\end{document}
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\end{document}
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@ -3,10 +3,11 @@
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\usepackage{nccmath}
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\usepackage{nccmath}
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\DeclareMathSizes{10}{10}{10}{10}
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\DeclareMathSizes{10}{10}{10}{10}
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\setlength{\parindent}{0pt}
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\setlength{\parindent}{0pt}
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\title{Struckturelle Induktion}
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\title{Strukturelle Induktion}
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\date{ }
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\date{ }
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\begin{document}
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\begin{document}
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\maketitle
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\section*{Strukturelle Induktion}
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%\maketitle
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%----------------------------------%
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%----------------------------------%
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\textbf{Referenzaufgabe:} \\
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\textbf{Referenzaufgabe:} \\
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\\
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\\
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@ -140,7 +140,7 @@
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- (1) $\rightarrow_{i}$\\
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- (1) $\rightarrow_{i}$\\
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- (2)(3) $\rightarrow_e$\\
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- (2)(3) $\rightarrow_e$\\
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- in (3) ist $[\,one:\mathrm{N}\,]$ mit Ax fertig\\
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- in (3) ist $[\,one:\mathrm{N}\,]$ mit Ax fertig\\
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- in (4) ???\\
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- in (4) $\forall _e$\\
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- (4)(links) $\forall_{e}$ (rechts) $\rightarrow_e$\\
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- (4)(links) $\forall_{e}$ (rechts) $\rightarrow_e$\\
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\end{small}
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\end{small}
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\\
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\\
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