0 votes 0 votes If L1 and L2 are Turing-Recognizable then L1 ∪ L2 will be decidable??? can i say L1 and L2 is re and re is closed under union operation so it is decidable?? abhishekmehta4u asked Apr 20, 2018 abhishekmehta4u 296 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply srestha commented Apr 20, 2018 reply Follow Share $L_{1}=RE$ If $L_{2}=RE$and complement of $L_{1}$ then only it will be decidable 0 votes 0 votes Soumya29 commented Apr 20, 2018 reply Follow Share RE $\cup$ RE = RE If $\bar{L1}$ and $\bar{L2}$ are also RE then only their union is Decidable i.e Recursive 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes I THINK NO BECAUSE DECIDABLE MEANS total turing machine i.e REC AND REL IS SEMIDECIDABLE BECAUSE IS RECOGNISED BY TURING MACHNIE NOT BY TOTAL TURING MACHINE eyeamgj answered Apr 20, 2018 eyeamgj comment Share Follow See all 0 reply Please log in or register to add a comment.