2,585 views
0 votes
0 votes
(A) Regular      (B)CFG

(C)Deterministic CFG    (D) Context Sensitive

2 Answers

Best answer
10 votes
10 votes

L1 - L2 = L1 ⋂ L2'

Now CFL complement need not be CFL (as CFL's are not closed under complement). But any CFL is also a CSL and CSLs are closed under complement. So, any CFL complement is guaranteed to be a CSL. Now, CSL intersection CSL (regular is also a CSL) is CSL, and so L1-L2 is always Context Sensitive. 

(It can also be regular or CFL but we can say always only with respect to CSL)

http://gatecse.in/wiki/Closure_Property_of_Language_Families

selected by
0 votes
0 votes
lower - higher = complement (higher)

and complement of CFG is not close so it is CSL

Related questions

5 votes
5 votes
4 answers
1
Purple asked Jan 28, 2016
8,961 views
Consider L1, L2 ⊆ Ʃ* such that L1 and L1 ∪ L2 are regular.(a) L2 is definitely regular(b) L2 may not be regular(c) L2 is context free(d) None of aboveIs it option B ...
0 votes
0 votes
0 answers
2
Sandeep Verma asked Nov 13, 2017
348 views
If a language(L1) is Recursive (REC) and L2 is Recursive (REC) , then what will be L1-L2 ?
3 votes
3 votes
2 answers
3
Sandeep Verma asked Nov 12, 2017
2,360 views
If a language(L1) is Recursive enumerable (RE) and L2 is Recursive (REC) , then what will be L1 - L2 ? Can we directly use set difference property or do the s...