Final answer:
Using the Pumping Lemma for context-free languages, we prove that the language L = n ≥ 0 is not context-free by showing that no division of a string in this language can satisfy the conditions of the lemma without disrupting the pattern of characters in L, resulting in a string not in L.
Step-by-step explanation:
Proving the language is not context-free using the Pumping Lemma
To prove that the given language L = 02n13n0n is not context-free, we use the Pumping Lemma for context-free languages. According to the Pumping Lemma, for any context-free language, there exists a number p (pumping length) such that any string s in L with length at least p can be divided into five parts, s = uvxyz, satisfying the conditions:
- For each i ≥ 0, the string u(vi)xy(zi) is also in L.
- |vxy| ≤ p.
- |vy| > 0.
Assume L is context-free and let p be the pumping length given by the Pumping Lemma. Consider the string s = 02p13p0p, which is in L and has a length greater than p. We attempt to divide s into five parts to satisfy the conditions of the Pumping Lemma. However, due to the structure of s (specific lengths of 0s and 1s), pumping any part of the string that includes both 0s and 1s will disrupt the required pattern of 02n13n0n, leading to a string that is not in L. This contradiction implies that our assumption that L is context-free is false, and therefore, L is not a context-free language.