Closure Properties of CFL’s

Theorem. The CFLs are closed under union, concatenation, and Kleene closure.
Theorem. If $A$ is a CFL and $B$ is a regular language, then $A\cap B$ is a CFL.
In the rest of this handout let $\Sigma=\{a, b, c\}$.
Lemma. The language $E = \{ w \in\Sigma^*: |w|_a = |w|_b = |w|_c \}$ is not a CFL.
Proof. Suppose it were. Then $F = E\cap L(a^*b^*c^*)$ would be a CFL. But $F = \{a^nb^nc^n : n\ge0\}$ and we have already seen that $F$ is not a CFL, a contradiction. \[\tag*{$\Box$}\]

More Theorems

Theorem. The CFLs are not closed under intersection.
Proof. This is because $E$ is the intersection of 2 CFLs: \[ E = \{\,w\in\Sigma^*:|w|_a = |w|_b\,\} \cap \{\,w\in\Sigma^*: |w|_b = |w|_c\,\}. \] To show the RHS languages are CFL’s, we can use either PDAs or CFGs. \[\tag*{$\Box$}\]
Theorem. The CFLs are not closed under complement.
Proof. $E$ is the complement of the CFL \[ \overline{E} = \{ w\in\Sigma^*: |w|_a \ne |w|_b\,\} \cup \{\, w\in\Sigma^*: |w|_b \ne |w|_c\}. \] To show the RHS languages are CFL’s, note that $i\ne j$ means $i<j$ or $i>j$. \[\tag*{$\Box$}\]
Another way to prove the above theorem is to note that the language \[\verb|SQ| = \{ww : w \in\{a,b\}^*\}\] is not context-free; however, its complement is.