B = {anbmcn | m, n ≥ 0}
Prove that the following language is not regular using the pumping lemma.
To prove that the language B = {a^nb^mc^n | m, n > 0} is not regular. To do this, we use the pumping lemma for regular languages, which is a powerful tool for proving that a language is not regular.
Here are the steps of the proof in more detail:
1. Assume that B is a regular language and let p be the pumping length. The pumping length is a constant that is guaranteed to exist if B is regular.
2. Consider the string w = a^pb^pc^p. This string is in B and has length |w| = 3p ≥ p. Since w is a member of B and its length is greater than or equal to the pumping length p, we can apply the pumping lemma to w.
3. According to the pumping lemma, we can write w as w = xyz, where |xy| ≤ p and |y| > 0.
This means that the string y contains at least one symbol from the set {a}. We can write x as a^k and y as a^l (where 1 ≤ l ≤ p). We can then write z as a^(p-k-l)b^pc^p.
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