We can recognize strings of given language using one Stack. These given languages are context free, so also context sensitive. Because CFLs are closed under Union property, so union of given languages will also be Context free. But CFLs are not closed under Intersection property, so Intersection of two CFLs may not be CFL. Given L1 and If L2 are two context free languages, their intersection L1 ? L2 is not context free because we cannot identify strings of resultant language with help of one Stack: L1 = { anbncm | n > 0 and m > 0 } and L2 = { ambncn | n > 0 and m > 0 } L3 = L1 ? L2 = { anbncn | n > 0 } is not context free. So, option (a) is false. See wikipedia page for closure properties.
Given a Turing Machine M = ({q0, q1, q2, q3}, {a, b}, {a, b, B}, ?, B, {q3}) Where ? is a transition function defined as ?(q0, a) = (q1, a, R) ?(q1, b) = (q2, b, R) ?(q2, a) = (q2, a, R) ?(q3, b) = (q3, b, R) The language L(M) accepted by the Turing Machine is given as:
Production Rule: aAb->agb belongs to which of the following category?
Correct Answer: C
Context Sensitive Language or Type 1 or Linearly Bounded Non deterministic Language has the production rule where the production is context dependent i.e. aAb->agb.