Similarly one may ask, is ETM Turing recognizable?
Since ETM is Turing-recognizable, this means that ATM is also Turing-recognizable (using Theorem 5.2 in the text), a contradiction (to Corollary 4.17). Alternatively, we could use Corollary 5.23 to derive a contradiction to (a). (b) L2 = {(M) | M is a Turing machine that halts on an empty input}.
Subsequently, question is, is ATM recognizable? Because we know that ATM is recognizable, our theorem implies that ATM and ATM are both decidable. But we know that ATM is not decidable. This is a contradiction, hence ATM cannot be recognizable. The language ATM and its undecidability (including proof).
Additionally, what is co Turing recognizable?
Intuitively, if a language is co-Turing-recognizable, it means that there is a computer program that, given a string not in the language, will eventually confirm that the string is not in the language. It might loop infinitely if the string is indeed within the language, though.
Is EQTM Decidable?
Theorem: EQTM is undecidable. We are getting tired of reducing A TM to everything. Let's try instead a reduction from EMPTY TM to EQTM.
