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You are confusing the langauges here $h(\langle M, w, \varepsilon\rangle) \notin L$ certainly, but this does not mean anything releated to $L_{halt}$. Assume $M$ halts on $w$ after $i$ moves. Then $\l …

Yes a TM also has states. Formally, one can define TMs as $(Q, \Sigma, \Gamma, \square, \delta, q_0, \bar{q})$ with states $Q$, input alphabet $\Sigma$ not including the blank symbol $\square$, tape a …

Before I actually start answering your question, I will prove the following (see my comment): A language $A \subseteq \{0, 1\}^\ast$ is recursively enumerable iff there exists a recursive langua …

First of all, note that there are undecidable languages where Rice's theorem cannot be applied. In your third paragraph one may suggest that it would work that way. But in fact, for $E_{TM}$ you can a …

First, note that if $|\Sigma| > 1$, you have more than eight different $w$. Second, I would assume that $M_w[\varepsilon]$ is the output of the Turing machine corresponding to encoding $w$ on input $\ …

If you have $g$ symbols (including a blank) and a tape of size $n$ then there are $g^n$ words of length $n$. This is really basic combinatorics: The reasoning is that you have $g$ options for the firs …

A set $A$ is computable (like in Turing machines) iff its characteristic function $$\chi_A(x) = \begin{cases}1, & x \in A\\ 0, & x \notin A\end{cases}$$ is recursive. The class of recursive functions …

Let $x = \langle M, w \rangle \in H$, where $H$ is the usual haltingproblem. Then $\langle M_2 \rangle$ halts on every input, especially on $\varepsilon$ and by construction accepts every input, espec …

First note, that for a regular language, however encoded, it is decidable wether it is finite or not. So you can use the language $REGU$ as an oracle for $FIN$ in the following way: To check if $\lang …

I think that "limits of computation" is a good entry point. Usually, computer scientists learn how to solve problems in an logical and algorithmic way. The whole concept of undecidability is to show t …

For $A = \{\langle G \rangle \mid L(G) \neq \Sigma^\ast\}$ this procedure, returns "yes" iff there is a word $w \notin L(G)$ for a given grammar encoding $\langle G \rangle$ and never halts otherwise …

If you consider a grammar $(N, \Sigma, P, S)$ you could actually give a regular expression for the production rules $P$, which pretty much determines the whole grammar: $$r = S \to (\bar{N} + \bar{\Si …

A linear bounded automaton is a Turing machine that runs on input of size $n$ in $\mathcal{O}(n)$ space. By the space hierachy theorem there exist languages that need e.g. $\omega(n^2)$ space.

If the task is to determine whether a given machine $M$ halts, when you already know that $M$s language is decidable then the answer is always "yes" because every TM that decides a language halts on e …

Using a fixed alphabet $\Sigma$, one can define regular expressions over $\Sigma$ inductively: $\varnothing$ and each $a \in \Sigma$ are regular expressions. if $r$ and $e$ are regular expressions th …