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[TI] Compact: Almost complete

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Janis Hutz
2025-12-09 10:58:20 +01:00
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semester3/ti-compact/parts/04_computability.tex
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@@ -39,9 +39,9 @@ Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines,
This however leads to a contradiction, as $w_i \in L_\text{diag} \Longleftrightarrow d_{ii} = 0 \Longleftrightarrow w_i \notin L(M_i)$.
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 $L_\text{diag} \in \cL_{RE}$
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 $L_\text{diag} \in \cL_{RE}$.
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 $w_i \notin L(M_i)$
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 $w_i \notin L(M_i)$.
\setLabelNumber{theorem}{3}
\inlinetheorem $L_\text{diag} \notin \cL_{RE}$
@@ -50,7 +50,12 @@ In other, more different, words, $w_i$ being in $L_\text{diag}$ implies (from th
\subsection{Reductions}
This is the start of the topics that are part of the endterm.
\label{sec:reductions}
This is the start of the topics that are explicitly part of the endterm.
For a language to be in $\cL_R$, in contrast to $L \in \cL_{RE}$, the TM has to halt also for \texttt{no} instances, i.e. it has to be an algorithm.
In other words: A TM $A$ can enumerate all valid strings of a \textit{recursively enumerable language} ($L \in \cL_{RE}$),
where for \textit{recursive languages}, it has to be able to difinitively answer for both \texttt{yes} and \texttt{no} and thus halt in finite time for both.
First off, a list of important languages for this and the next section:
\begin{itemize}