Is this language Context-Free? Possible Duplicate:
  Is this language Context-Free?  




The language is defined by
$$(a+b)^*-\{(a^nb^n)^n\mid n \geq1 \}$$

is Context-Free Language? I believe that the answer is that it is not a CFL's, but I can't prove it by Ogden lemma or pumping lemma.

chazisop, (a+b)* means {a,b}*. That's all strings that you can form with the symbols a or b or both.
If I understand the notation correctly, only $ab$ can fit both $(a+b)^{*}$ and $(a^{n}b^{n})^{n}$. I see no other way of stacking more $b$'s after the a's, since $a+$ suggest you should have at least one $a$ in every repetition of the expression in parentheses.

$(a+b)^{*}-ab$ seems like a regular language to me, since $ab$ rejects, $a+b$ just consumes one or more $a$'s before a $b$. If we can construct a DFA for $a+b$, we can construct one for $(a+b)^{*}$ as well.