Answer:
<em>L is not a regular language with formal proofs </em>
Explanation:
<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages,
</em>
<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p,
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<em>s = xyz subject to the following conditions:
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<em>(a) |y| > 0
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<em>(b) |xy| ≤ p, and
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<em>(c) ∀i > 0, xyi
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<em>z ∈ L</em>
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<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>
<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows :
</em>
<em>(a) |y| > 0
</em>
<em>(b) |xy| ≤ p, and
</em>
<em>(c) ∀i > 0, xyi
</em>
<em>z ∈ L.
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<em>Choose s = 0p10p
</em>
<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>
<em>for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0
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<em>z should be in L. xy0
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<em>z = xz = 0(p−k)10p
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<em>It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language</em>
and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
