0 votes 0 votes From http://www.cs.rice.edu/~nakhleh/COMP481/final_review_sp06_sol.pdf $L_{26}=\{<M>|$ M is a TM such that both L(M) and $\lnot L(M)$ are infinite $\}$ I was unable to get proof given in pdf above.Can anyone explain, if someone got it. Theory of Computation decidability theory-of-computation + – Ayush Upadhyaya asked Jan 13, 2019 Ayush Upadhyaya 437 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Shobhit Joshi commented Jan 13, 2019 reply Follow Share Say, $M_1=\left \{a^n:n\,\,is\,\,a\,\,multiple\,\,of\,\,2\right \}$ $\overline{M_1}$ will be strings where n will not be multiple of 2 So, both $M_1$ and $\overline{M_1}$ are infinite. So, $L(M_1)=T_{yes}$ $M_2=\left \{a^n:n\geq 0\text{}\right \}$ $\overline{M_2}=\phi$ So, $M_1$ is infinite but $\overline{M_1}$ is not. So, $L(M_2)=T_{no}$ So, $T_{yes}\subset T_{no}$ as $M_1\subset M_2$ So, it is a non-monotonous property. Hence non-RE Correct me if I'm wrong 1 votes 1 votes Ayush Upadhyaya commented Jan 14, 2019 reply Follow Share @Shobhit Joshi-Seems good to me :).But I can't detect any flaw here so be cautious. 0 votes 0 votes Shobhit Joshi commented Jan 14, 2019 reply Follow Share Works almost all the time, let's see if it works in gate ! 1 votes 1 votes Ayush Upadhyaya commented Jan 14, 2019 reply Follow Share Can I take the machine M2 to be the machine that accepts everything? 0 votes 0 votes Shobhit Joshi commented Jan 14, 2019 reply Follow Share it accepts every string of length >= 0, using only $a$ 0 votes 0 votes Please log in or register to add a comment.