Consider the language :

L = {a^i b^j c^k | i, j, k >= 0 and (i != j or j != k)}

Prove that this language is context-free by giving a grammar but not deterministic context-

free by showing that its complement is not context-free. The grammar can be rather long

to write out fully so it suces to explain what you are doing and reduce it to similar cases

and then say, "this case is just like what we have done before." Of course, if you want to

write out all the rules you are free to do so. Some of you might nd that easier than writing

explanations. To prove that the complement is not context free can be done by a reduction

argument to a familiar language that is known to be not context free.