(Non) Context Free language? In an exam question it is asked to identify if the given language is Context-Free, Non Context Free or Regular.
This is the question :



In my opinion this language is not a CFL since w doesn't belong to L', meaning that it belongs to not(L') which is all words (a^n b^n c^n) which is not CFL.

Apparently my answer is not right and I really don't understand why... Is this question a "trap" because i,j,k > 0 and not >= ? Meaning that a word (b^n c^n) is in L ? Looks like I'm lost ! 

I really need your help and explantions.

Thanks in advance.
It helps to think of $L$ as having two parts.


First part: $L_1$ is all strings not in the form $a^i b^j c^k$. We can write this with the regular expression $\neg [a^* b^* c^*]$, so it's regular.


(I've been informed this notation isn't universal, so to spell it out: it's the complement of the regular language $a^* b^* c^*$, and the complement of a regular language is regular.)

Second part: $L_2$ is all strings in the form $a^i b^j c^k$ such that the given conditions on $i, j, k$ are not fulfilled.


The given conditions are that $i, j, k$ are all different. So how can those conditions be violated? Either $i=j$, or $j=k$, or $i=k$. We can call these three options $L_{2a}$, $L_{2b}$, and $L_{2c}$ if you want.

Let's take just one of them as an example; the others work the same way. $L_{2a}$ is all strings of the form $a^i b^i c^k$. Can you write a regular expression for this? How about a CFG? (The answers are no, and yes: in fact, this is one of the classic examples for what a CFG can do that a regular expression can't.)

Now, since $L_{2a}$, $L_{2b}$, and $L_{2c}$ are all context-free, their union ($L_2$) is also context-free. And since $L_1$ is regular, it's also context-free. So $L$, as the union of two context-free languages, is context-free.