Closure of CFL

Closure properties of CFL.

Recall definition of "closure":

   A set, S, is closed under a unary operation, 'O',
   if given x in S, the result of O(x) is also in S.

E.g., the integers is closed under unary minus:

   x in Z then so is -x

The set of positive integers is not closed under '-'.


   A set, S, is closed under a binary operation, 'O',
   if given x, y in S, the result of x O y is also in S.


[Give examples which are closed.]

[Give examples which are not closed.]

Finally, generically:

   A set, S, is closed under an operation, 'O', if the
   result of applying 'O' to members of the set yields
   a member of the set.



Theorem:

Context Free Languages are closed under:

   union
   concatenation
   Kleene star


Note - the text constructs an "effective proof" which is
worth knowing:

  i) Union: add productions S -> S1 | S2

 ii) Concat.: add production S -> S1 S2

iii) K.S.: add productions S -> S1 S | lambda



Theorem:

The set of CFL is not closed under intersection or complementation.

Note - the text constructs a proof "by counter example" which is
worth noting:

                         i i j
 i) intersection: L  = {a b c | i, j >= 0}
                   1
                         j i i
                  L  = {a b c | i, j >= 0}
                   2


    G : S -> BC
     1  B -> aBb
             | lambda
        C -> cC
             | lambda

    G : S -> AB
     2  A -> aA
             | lambda
        B -> bBc
             | lambda


[good examples to ask: what is the language L(G)?]


 since the intersection of L  and L is
                            1      2

                         i i i
                  L  = {a b c | i >= 0}

 and this has been shown to not be CF, it must be that CFL are
 not closed under intersection.




ii) Complementation:

The text's proof is a good use of DeMorgan's Laws and

   "proof by contradiction":

   Assume that the set of CFL is closed under complementation.
   Let L1 and L2 be any two CFL.


                      _______
                      _     _
   By hypothesis, L = L  U  L   = L  Intersect L
                       1     2     1            2

   is context-free. Thus the intersection of any two CFL's is
   a CFL, which contradicts part i.