Answer :
Final answer:
The language L1 = {w ∈ {a, b, c, d}* | |w|a ≥ |w|c} is not a regular language. We can prove this using the pumping lemma for regular languages.
Explanation:
The language L1 = {w ∈ {a, b, c, d}* | |w|a ≥ |w|c} is not a regular language.
To prove this, we can use the pumping lemma for regular languages. Let's assume that L1 is regular and generate a contradiction.
- Assume L1 is regular.
- Let p be the pumping length of L1.
- Choose the string w = a^p c^p ∈ L1.
- According to the pumping lemma, we can write w as xyz, where |xy| ≤ p, |y| > 0, and for all i ≥ 0, xy^iz ∈ L1.
- Now let's consider the case when x = a^k, y = a^l, and z = a^(p-k-l) c^p. Since xy^2z ∈ L1, we have |xy^2z|a ≥ |xy^2z|c.
- Simplifying this inequality, we have p + 2l ≥ p.
- However, this inequality is not possible since l > 0. Thus, we have a contradiction, which implies that L1 is not regular.
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