Given ,
$L1$ is decidable language/Recursive language
$L2$ is Recursive enumerable language(REL)/Turing recognizable language
Some facts:-
- Recursive language is also recursive enumerable language.
- Recursive language is closed under complementation ,intersection ,union ,etc.
- Recursive enumerable language is closed under union,intersection but not closed under complementation.
- Complement of recursive enumerable language can be REL or NON REL as well .
Option 1:-
$\bar{L_{1}}\cap \bar{L_{2}}$ =$\bar{\left ( {L_{1}}\cup {L_{2}} \right )}$ [demorgan law]
Now as recursive language is also recursive enumerable language we can consider L1 also as Recursive enumerable language.
Now , $( {L_{1}}\cup {L_{2}})$ is union of two recursive enumerable language which also recursive enumerable .
now complement of ${{L_{1}}\cup {L_{2}}}$ is may be recursive enumerable or may not be as recursive enumerable language is not closed under complementation.
so ,$\bar{L_{1}}\cap \bar{L_{2}}$ is not turing recognizable language.
Option 2:-
$L_{1 }\cap L_{2}$
$L1$ is recursive language so recursive enumerable language as well.
$L2$ is recursive enumerable language.
So , $L_{1 }\cap L_{2}$ is also recursive enumerable language as recursive enumerable language is closed under complementation.
Now every recursive language is recursive enumerable language but not every recursive enumerable language is recursive .
So, $L_{1 }\cap L_{2}$ is not Turing decidable language.
Option 3:-
$\large \frac{L_{1}}{L_{2}}$ (I am considering the given operation as set difference operator)
=$\large L_{1}-L_{2}$
=$\large L_{1}\cap \bar{L_{2}}$
Now $L_{2}$ is recursive enumerable language , So,it’s complement is not closed as it may or may not be recursive enumerable after complementation.
So $\large L_{1}\cap \bar{L_{2}}$ is not turing decidable language.
Option 4:-
$\large \frac{L_{2}}{L_{1}}$
= $\large L_{2}-L_{1}$
= $\large L_{2}\cap \bar{L_{1}}$
Now $L_{1}$ is recursive language and complement of recursive language is recursive so ,$\bar{L_{1}}$ recursive so as well as recursive enumerable also.
$L_{2}$ is recursive enumerable language.
Now , Recursive enumerable language is closed under intersection So,$\large L_{2}\cap \bar{L_{1}}$ is recursive enumerable.
So,$\large \frac{L_{2}}{L_{1}}$ is turing recognizable language.
Correct answer is (D).