Finalized proof to theorem 2
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@ -326,10 +326,19 @@ analog zu \autoref{def:kaffeekassentransition}.
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Dann existieren laut \autoref{def:explizitekaffeekasse} \(\delta_{i,j} \in \integers\)
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für alle \(i,j \in \naturals_{\leq n}\) als Einträge in \(\kappa\). Setze nun
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\[ \Delta_i = \sum_{l=1}^{n} \delta_{i,l} . \]
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Dann gilt
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\[ \sum_ \]
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ergibt sich die bilanzierende Kaffeekasse als
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\( k = (\Delta_1,\ldots,\Delta_n) \)
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Trivialerweise gilt
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\[
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\begin{array}{lll}
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\sum\limits_{i=1}^n \Delta_i & = & \sum\limits_{i=1}^n \sum\limits_{l=1}^{n} \delta_{i,l} \\
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& = & \sum\limits_{i=1}^n \delta_{i,i} + \sum\limits_{i=1}^n \sum\limits_{l=1}^{i-1} \delta_{i,l} + \sum\limits_{i=1}^n \sum\limits_{l=i+1}^n \delta_{i,l} \\
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& = & \sum\limits_{i=1}^n \delta_{i,i} + \sum\limits_{i=1}^n \sum\limits_{l=1}^{i-1} \delta_{i,l} + \sum\limits_{i=1}^n \sum\limits_{l=1}^{i-1} - \delta_{i,l} \\
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& = & \sum\limits_{i=1}^n \delta_{i,i} + \sum\limits_{i=1}^n \sum\limits_{l=1}^{i-1} (\delta_{i,l} - \delta_{i,l}) \\
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& = & 0.
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\end{array}
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\]
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Und die bilanzierende Kaffeekasse
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ergibt sich also als
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\( k = (\Delta_1,\ldots,\Delta_n) \).
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\end{proof}
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\begin{beob}[Kaffeeparadoxon]
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