Consider the following language:

L1 = {w ∈ {a, b, c, d}* | |w|a ≥ |w|c}

Is L1 a regular language? Prove your answer by either developing a DFA for L1 or using the pumping lemma to show non-regularity.

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High School

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smith4567 smith4567 · Answer 1 · 2023-07-19T01:05:48-04:00

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.

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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