theory of computation

Question

Let C be a context-free language and R be a regular language. Prove that
the language C $\cap$ R is context free.

Answer 1

A language is context-free only when it has only one condition in the formation of language.

For example $ L_1 = \{ a^n b^n | n>= 0 \} $

A regular language, will not have any conditions.

For example $ L_2 = {(a + b)}^* $

So Intersection, of condition and non condition will return that one condition,

Hence $ L_1 \cap L_2 = L_1 = \{ a^n b^n | n>= 0 \} $ Hence, Context free.

Comments

L1 U L2 will it be DCFL or NCFL or not CFL ?? and how ??

Please explain

Answer 2

C: Context free language

R: Regular language

Every regular is context free but every context free is not regular

C(Context free)  INTERSECT  R(Regular)

C(Context free)  INTERSECT   R(Context free)                           Every Regular is context free

Output: CFL is not closed under Intersection

how can this be true  because CFl are not closed under intersection!!!!

plz make me correct if  i m wrong

Comments

CFL is not closed under intersection doesn't mean for Any  CFL's intersection can't be CFL

for some CFL's it can't be CFL but not for all

This is what i think