0 0 votes $L=\left \{\langle M_{1},M_{2}\rangle \text{ such that L}(M_{1})\prec L(M_{2}) \right \}$ is it recursive enumerable? here $L\left ( M_{1} \right )\prec L\left ( M_{2} \right )$ signifies language $L\left ( M_{1} \right )$ is reducible to $L\left ( M_{2} \right )$ Theory of Computation theory-of-computation turing-machine decidability recursive-and-recursively-enumerable-languages + – ajaysoni1924 1.5k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes Last time I said as it was undecidable it is actually Semi Decidable. We can't really say. As the condition M1 <M2 so Semi Decidability is satisfied. Hence, M1 is reduciblie to M2 is recursive enumerable. One way theorem satisfies these conditions of semi-decidability are: 1. P <--- P 2. NP <---- NP 3. Recursive Enumerable <---- Recursive enumerable 4. Recursive lang <--- Recursive Language hina firdaus answered Jul 16, 2019 • edited Jul 19, 2019 by hina firdaus hina firdaus comment Share Follow See all 4 Comments 4 4 Comments reply Arjun commented Jul 16, 2019 reply Follow flag Didn't understand anything 😞 1 1 replyShare ajaysoni1924 commented Jul 17, 2019 reply Follow flag @Arjun Sir, is not this question is incomplete? as we have not given any description about machine M either it is decidable or undecidable than how we can decide, in that case it should be not even RE. not getting answer provided by@hina firdaus 0 0 replyShare Arjun commented Jul 17, 2019 reply Follow flag M1 and M2 are inputs or like parameters to a function. That is from the input you can know the description of the 2 Turing machines and based on this you have to say if first one can be reduced to second or not -- so question is perfectly fine. For some inputs this is easy to say yes or no as answer but question is if we can always say yes or no correctly for all inputs. 3 3 replyShare gajendercse commented Jan 4, 2021 reply Follow flag so it is semidecidable or non R.E? 0 0 replyShare Please log in or register to add a comment.