Final answer:
The language {a^n b^n c^n: n ≥ 0} is not context-free because a context-free grammar cannot generate strings where the number of a's, b's, and c's are all equal and must occur in sequence for some non-negative integer n. The statement is false.
Step-by-step explanation:
The statement in question is regarding whether the language {aⁿ bⁿ cⁿ : n ≥ 0} is context-free. A context-free language is one that can be generated by a context-free grammar (CFG), which has rules that allow a symbol to be replaced by a sequence of symbols, independent of the surrounding symbols (context).
This particular language represents strings where the number of a's, b's, and c's are all equal and must occur in the sequence of aaa...bbb...ccc for some non-negative integer n. However, it is well-known in formal language theory that this language is not context-free.
The reason is that CFGs lack the ability to count and ensure that the number of a's, b's, and c's are all equal, especially when the number of each can be arbitrarily large. This is evidenced by the pumping lemma for context-free languages, which this language does not satisfy.