# Is language $a^mb^nc^n, m \not= n$ context free

I need to say Is language $$a^mb^nc^n, m \not= n$$ context free

I managed to find a grammar for $$L1 =$$ { $$a^lb^mc^n | l=m$$ or $$m = n$$ }, but I couldn't find the one I needed. Maybe it is impossible, but than why? If it may be helpful, that is the grammar for L1 wich i found: $$A \rightarrow \epsilon |aAb, B \rightarrow \epsilon|bBc, S1 \rightarrow A|S1c, S2 \rightarrow B|aS2, S\rightarrow S1|S2$$

I think that is not context free at all but I don't know how to proof.

The reason that $$L_1$$ is context-free is that one needs to keep track of only two numbers at the same time: either $$l=m$$ or $$m=n$$. The 'or' is handled using union, we patch two languages together. The difference of the two numbers $$l-m$$ or $$m-n$$ can be stored on the stack while reading the string.
Your language $$L=\{a^mb^nc^n \mid m\neq n\}$$ is not context-free. Intuitively because we need to handle three numbers, and there is no way to store two differences.
The formal proof of non-context-freeness can be done using the pumping lemma for context-free languages. As the specification involves the unequality of $$m$$ and $$n$$ a specific trick is needed, which involves a factorial. An example of this trick is shown here for nonregularity: Prove if $$L=\{0^m1^n∣m≠n\}$$ is regular or not. The approach is basically the same.