Week8 monday

Theorem: \(A_{TM}\) is not Turing-decidable.

Proof: Suppose towards a contradiction that there is a Turing machine that decides \(A_{TM}\). We call this presumed machine \(M_{ATM}\).

By assumption, for every Turing machine \(M\) and every string \(w\)

Define a new Turing machine using the high-level description:

\(D =\)“ On input \(\langle M \rangle\), where \(M\) is a Turing machine:

Is \(D\) a Turing machine?

Is \(D\) a decider?

What is the result of the computation of \(D\) on \(\langle D \rangle\)?

Theorem (Sipser Theorem 4.22): A language is Turing-decidable if and only if both it and its complement are Turing-recognizable.

Proof, first direction: Suppose language \(L\) is Turing-decidable. WTS that both it and its complement are Turing-recognizable.

Proof, second direction: Suppose language \(L\) is Turing-recognizable, and so is its complement. WTS that \(L\) is Turing-decidable.

Give an example of a decidable set:

Give an example of a recognizable undecidable set:

Give an example of an unrecognizable set:

True or False: The class of Turing-decidable languages is closed under complementation?

Definition: A language \(L\) over an alphabet \(\Sigma\) is called co-recognizable if its complement, defined as \(\Sigma^* \setminus L = \{ x \in \Sigma^* \mid x \notin L \}\), is Turing-recognizable.

Notation: The complement of a set \(X\) is denoted with a superscript \(c\), \(X^c\), or an overline, \(\overline{X}\).