Final answer:
Using the Pumping Lemma for CFLs, by assuming a language L where the number of a's, b's, and c's are equal is context-free and showing that for any string of sufficient length, we cannot satisfy the lemma's conditions, we conclude that L is not a context-free language.
Step-by-step explanation:
To show that a given language is not context-free using the Pumping Lemma for CFLs, we need to assume that the language is context-free, and therefore there exists some pumping length p such that any string w in the language of length at least p can be divided into five parts, uvxyz, satisfying the conditions of the pumping lemma for CFLs. Let's consider the language L = {w ∈ {a,b,c}∗: n_a(w) = n_b(w) = n_c(w)}. Assume that L is context-free and let p be the pumping length. Consider the string s = a^p b^p c^p, which is in L and has a length greater than p. According to the pumping lemma, s can be split into five parts, uvxyz, such that these conditions hold:For each i ≥ 0, the string uv^ixy^iz is in L. |vy| > 0, |vxy| ≤ p
Since |vxy| is less than or equal to p, the pumped strings will have an imbalance in the number of a's, b's, and c's when we pick values of i other than 1. Thus, these strings cannot be in L, contradicting the assumption that L is context-free. Therefore, L is not a context-free language.