is etm recognizable

Skip ahead to section 2.2 or 2.3 and show that ETM is not recognizable. Then by the theorem that. “a language is decidable iff it is both recognizable and 

This in turn implies that ETM is undecidable. At least one of them must be not recognizable. ETM is recognizable. ETM is not recognizable.

Consider the emptiness problem for Turing machines: ETM = { ?M?

However ETM is Turing-recognizable (HW 8

If both L and L are recognizable then L is decidable. As we know that ATM HALTTM

But ¬ETM is Turing recognizable and ¬ATM is not

HALTTM is Turing-recognizable since it can be recognized by TM U. HALTTM is not Turing-decidable. ETM = {< M >

Since ETM is undecidable (Theorem 5.2) EQTM must be undecidable. • We'll see later that EQTM is not Turing-recognizable not co-Turing-recognizable 

Show that the language ETM = {?M?

4 Mar 2019 But ETM is Turing-recognizable (why?) which would mean ATM recognizable (false). ? Mapping reductions may not exist!

Skip ahead to section 2 2 or 2 3 and show that ETM is not recognizable Then by the theorem that “a language is decidable iff it is both recognizable and 

However ETM is Turing-recognizable (HW 8 problem 4) and ATM is not Turing-recognizable (Corollary 4 23) contradicting Theorem 5 22 3 Consider the language

Rice's Theorem: any nontrivial property of the language of a TM is undecidable • ETM is not Turing-recognizable • EQTM is neither Turing-recognizable nor co- 

Show that the language ETM = {?M?M is a Turing machines and L(M) = ? } is not Turing-recognizable Proof: We provide two different proofs Proof 1: We 

As we know that ATM HALTTM ETM are recognizable their complements cannot be rec- ognizable because then the languages would be decidable and we know that 

HALTTM is Turing-recognizable since it can be recognized by TM U HALTTM is not Turing-decidable Proof: We will reduce ATM to HALTTM Assume TM R decides 

HALTTM = {?Mw? M is a TM that halts on input w} is undecidable If A ?m B and B is recognizable then A is recognizable Corollary 5 23

26 oct 2020 · Theorem: Every nontrivial property about recognizable languages (of Turing machines) is undecidable • The proof is a generalization of the 

29 juil 2013 · We mentioned that ETM is co-TM recognizable We will prove next that ETM is undecidable Intuition: You cannot solve this problem UNLESS

HALTTM is undecidable since ATM is undecidable This in turn implies that ETM is undecidable At least one of them must be not recognizable