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Let [tex]L_1[/tex] be the language [tex]L_1 = \{a^i b^p \mid i \geq 0 \text{ and } p \text{ is a prime}\}[/tex].

Define [tex]L_2 = b^* \cup L_1[/tex].

1. Prove that the language [tex]L_2[/tex] is not regular.

Answer :

Final answer:

L2 is not a regular language because it fails the Pumping Lemma for regular languages. An example string that repeats the middle part b two times results in a length not in L2, thus proving the language is not regular.

Explanation:

According to the Pumping Lemma for regular languages, if L2 is a regular language, there must be an integer p such that any string w in L2 of length at least p can be divided into three parts, x, y, and z with the properties of pumping lemma. However, based on the definition of L2, it contains strings of the form a^i b^p where p is a prime number. Select such a string where i = 0 and repeat the middle part b two times, we get b^(2p), which is not a prime number and thus not in L2. As Pumping Lemma doesn't hold for L2, it is not regular.

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