Question

-. If L and L2 are regular languages, show the the language BothOr Neither is also regular. Both Or Neither is the language that contains strings that are in both L1 and L, or in neither L or L2.

Answer 1

First we have to prove that if L, in regular, Lo io regular than LULZ is also regular. Ao L. La are regular over Zi, there must exist FA M = (8., Z, S. 2o. F) and M2 = (22 Z, dz. 102 F) such that LEM, and La E Ma. Assume that there is no common state between s, and Q2, i.e., 8, 082=0. Define another FA, M₂ = (8,{, d, V, F) where 1. = 8, UQ2 U f qo} , where qo is the new state & 8,082 2. F= F, OF 3. Transitional function s is defined as 8(90, E) - {woj, voz} 8(9. E) ~d, (9,6) if QE QI Sca,E) - h (92) if qt Q2 It is clear from the previous discussion that from To we can reach either the initial state a of My on the initial state of M₂ , i.e., q. Tramitions for the new FA, M, are similar to the transition of My and Ma.. A FF, UF, any shing accepted by M, or M₂ will also be accepted by M. Therefore, L, U is also regular.

First we have to prove complement of a regular language is abo regular. of Lid regular, we have to prove that I is also regular. AL is regular, there must be an FA, M=(0, Z, SVF) according L. Mis an FA, and so M has a traditional system Let us construct another traditional system M' with the Slate diagram of but reversing the direction of directed edges. Mocan be designed as follows: 4. The set of states of M' in the same asm. 2. The set of input symbols of m' in the same as M. 3. The initial state of m' is the same as the final state of m (M' is the reverse dineation of m) 4. The final state of M' in the same as the initial state of M CM is the reverse direction of m) let a sthing u belong to Line, WEM. So there is a path from vo to with path value W. By reversing the edges we get a path from F to go (beginning and final state of my am M. The path value is the reeverse of w ie, W. so, WEM'' So complement of a RL is also a RL.

proving , in regulars. From D' Morgan's theorem, we know. 2n2 = (1,6 u Lee We know, L., are regular. So, le 2 are also regular and union of two Rhs are also nagular. Then again complement of that language to regular. So, 4n in regulas.

or in neither regular a n na leo Rothor Neither so the language that contai that shings are are in in both both Li Li and and 2 L . L orla. L= (2,0%) u (2,022) Lin is proved, Lulzio regular. complement of RL is also regulas. L3=LOL. Lu=L, UL. LE=L, UL L3,L4, LG are all regular. so, L= L 3 U LE so, Lio regular. (proved).