Prove that if [tex]L_1[/tex] and [tex]L_2[/tex] are regular languages, then [tex]L_1[/tex] is equal to [tex]L_2[/tex] if and only if [tex]L_3 = (L_1 \setminus L_2) \cup (L_2 \setminus L_1)[/tex] is empty.

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Answer :

saumyasrivastavasyne saumyasrivastavasyne · Answer 1 · 2024-09-14T19:40:48-04:00

If Lị is equal to L2, then L3 = (L, NL ) u (Lin L2) is empty, and vice versa.

To prove that Lị is equal to L2 if and only if L3 = (L, NL ) u (Lin L2) is empty, we need to show both directions of the implication.

Direction 1: If Lị is equal to L2, then L3 is empty.

Assume Lị is equal to L2.
Since Lị is equal to L2, it means that they recognize the same set of strings.
Let's consider the language L3 = (L, NL ) u (Lin L2).
If L3 is not empty, it means there exists a string that is either in L or in NL, or in Lin L2.
Since Lị is equal to L2, any string in Lin L2 is also in Lị.
Therefore, if L3 is not empty, it implies that there exists a string in Lị that is either in L or in NL.
This contradicts the assumption that Lị is equal to L2, as it would mean that there exists a string in Lị that is not in L2.
Hence, if Lị is equal to L2, then L3 is empty.

Direction 2: If L3 is empty, then Lị is equal to L2.

Assume L3 is empty.
Let's consider the language Lị.
If Lị is not equal to L2, it means they recognize different sets of strings.
This implies that there exists a string in Lị that is not in L2, or a string in L2 that is not in Lị.
Since Lị is a regular language, it can be defined using a regular expression or a finite automaton.
By the definition of L3 = (L, NL ) u (Lin L2), if L3 is empty, it means that there does not exist a string that is either in L or in NL, or in Lin L2.
Therefore, if L3 is empty, it implies that there does not exist a string in Lị that is either in L or in NL.
This contradicts the assumption that there exists a string in Lị that is not in L2.
Hence, if L3 is empty, then Lị is equal to L2.

Therefore, we have proved that Lị is equal to L2 if and only if L3 = (L, NL ) u (Lin L2) is empty.

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