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48 lines
2.3 KiB
TeX
48 lines
2.3 KiB
TeX
\newsection
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\section{Computability}
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\stepcounter{subsection}
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\subsection{Diagonalization}
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The \bi{set of binary encodings of all TMs} is denoted $\text{KodTM}$ and $\text{KodTM} \subseteq \wordbool$ and the upper bound of the cardinality is $|\wordbool|$,
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as there are infinitely many TMs.
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Below is a list of countable objects. They all have corresponding Lemmas in the script, but omitted here:
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\drmvspace
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\begin{multicols}{4}
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\begin{itemize}
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\item $\word$ for any $\Sigma$
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\item $\text{KodTM}$
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\item $\N \times \N$
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\item $\Q^+$
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\end{itemize}
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\end{multicols}
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\rmvspace
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\drmvspace
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The following objects are uncountable: $[0, 1]$, $\R$, $\cP(\wordbool)$
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\inlinecorollary $|\text{KodTM}| < |\cP(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$
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\fhlc{Cyan}{Proof of $L$ (not) recursively enumerable}
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Proving that a language \textit{is} recursively enumerable is as difficult as providing a Turing Machine that accepts it.
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Proving that a language is \textit{not} recursively enumerable is likely easier. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$.
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\inlineex Assume towards contradiction that $L_\text{diag} \in \cL_{RE}$. Let
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\rmvspace
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\begin{align*}
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L_{\text{diag}} & = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and $M_i$ does not accept } w_i \} \\
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& = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and } d_{ii} = 0\}
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\end{align*}
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Thus assume that, $L_\text{diag} = L(M)$ for a Turing Machine $M$.
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Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines, so there exists an $i \in \N - \{ 0 \}$, such that $M = M_i$.
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This however leads to a contradiction, as $w_i \in L_\text{diag} \Longleftrightarrow d_{ii} = 0 \Longleftrightarrow w_i \notin L(M_i)$.
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In other words, $w_i$ is in $L_\text{diag}$ if and only if $w_i$ is not in $L(M_i)$, which contradicts our statement above, in which we assumed that $L_\text{diag} \in \cL_{RE}$
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In other, more different, words, $w_i$ being in $L_\text{diag}$ implies (from the definition) that $d_{ii} = 0$, which from its definition implies that $w_i \notin L(M_i)$
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\setLabelNumber{theorem}{3}
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\inlinetheorem $L_\text{diag} \notin \cL_{RE}$
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