Test: Identify Class Language

L is regular.

• Given L1 is a context free language and L2 as a regular language then L1-L2 is context free language and
• ~L1(complement of context free) is not context free because context free is not closed under complement.
• Complement of regular language i.e. ~L2 is also regular ( so d is correct ) since statements a and c are false.

Hence,Option (C) is correct.

Language generated by this grammar is L = (00)^+ that is regular but not 0^+

With only finite positions in stack, we can have only finite configurations and these can also be modeled as states in a finite
automata.

Language is not regular bcoz we need to match count of c's is equal to count of b's + count of a's

and that can implement by PDA

∂(q0,a,^)= (q0,a) [ push a in stack, as per language a comes first]

∂(q0,a,a)= (q0,aa) [push all a's into stack]
∂(q0,b,a) = (q1,ba) [push b in stack, state change to q1 that sure b comes after a]
∂(q1,b,b)=(q1,bb) [push all b's in stack]
∂(q1,c,b) = (q2,^) [ pop one b for one c]

∂(q2,c,b) = (q2,c) [ pop one b's for each c and continue same ]
∂(q2,c,a) = (q3,^) [ pop one a for one c , when there is no more b in stack]
∂(q3,c,a) = (q3,^) [pop one a for each c and continue same]
∂(q3,^,^) = (qf,^) [ if sum of c's is sum of a's and b's then stack will be empty when there is no c in input]

Note :1. state changes make sure b's comes after a and c's comes after b's]

L1 is CFL [ put a's in stack, and pop a with each b]]
L2 is CFL [put b's in stack and pop b with each c ]
c) is True.

b) is True CFL is closed under Union [ S-> S1 | S2 where S1 is grammar for L1 and S2 for L2]
CFL is not closed under Intersection, so intersection of two CFLs may or may not be CFL.

Let's examine:
L1 ∩ L2 = { aⁱbⁱcⁱ , i>0 } which is Context sensitive but not context free [can't match a's,b's and c's with one stack]
So a) is False

d) is True

Take L = Σ* then Lc = ∅ and Mc ∪ ∅ = M^c.

We know that complement of CFL need not be a CFL as CFL is not closed under complement

So, (A) and (B) are false.

If we take L = ∅ then L^c = Σ* and M^c ∪ Σ* = Σ* which is regular - (C) is also false.

So, answer (D)

Regular (in fact finite). Since n is finite, we have a finite set of strings satisfying the given conditions.

So, we can easily make a finite automata for those strings.

L1 = { s∈(0+1)* ∣ d(s) mod 5=2 } is regular having 2 as final state out of {0,1,2,3,4} states as given in example in posted link [in same DFA , final state will be 2 instead of 0 ]

Similarly L2 = { s∈(0+1)* ∣ d(s) mod 7≠4 } is also regular having states {0,1,2,3,4,5,6} and all are final state except 4
L1 ∩ L2 is L (given problem ) is also regular
As regular languages are closed under intersection.

Hence, D is the correct option.

A. True: DCFL are closed under Intersection with Regular Languages
C. True: L1 is regular hence also CFL and every DCFL is also CFL and All CFL are closed under Union.
D. True: L1 is regular hence also RE; L2 is DCFL hence also RE; RE languages are closed under Intersection
B. False: L1 is recursive hence also decidable, L3 is RE but not Recursive hence it is undecidable. Intersection of Recursive
language and Recursive Enumerable language is Recursive Enumerable language.