TOC#DECIDABLE

Question

L1= RE U NOT RE.

$L2=RE\cap NOT RE$.  WHERE RE IS recursively enumerable.

then L1 and L2 are ???

Comments

$L_{1}=\Sigma ^{*}$

$L_{2}=\phi$

Answer 1

l1 is set of all language over inputs BECAUSE it is beyond the turing recognizable 

nd l2 is empty lannguage becz RE∩NOTRE =RE-RE

Answer 2

Both are Regular Languages 
L1- RE U NOT RE = $\sum$^*  $\rightarrow$ Regular

L2- RE ∩ NOT RE = $\phi \rightarrow$ Regular