Answer :
An example of languages L1 and L2 is provided where neither L1 nor L2 is a subset of the other, but (L1∗ ∪ L2∗) = (L1 ∪ L2)∗. The example consists of L1 = {a} and L2 = {b}.
In this case, L1* includes all strings composed of multiple 'a's or the empty string, while L2* includes all strings composed of multiple 'b's or the empty string. The union of L1 and L2, (L1 ∪ L2), is the set {a, b}. The Kleene star operation applied to (L1 ∪ L2) gives the set of all possible combinations of 'a's and 'b's, which is also represented by (L1∗ ∪ L2∗).
Let's consider the example where L1 = {a} and L2 = {b}. In this case, L1* includes all strings composed of multiple 'a's or the empty string: L1* = {ε, a, aa, aaa, ...}. Similarly, L2* includes all strings composed of multiple 'b's or the empty string: L2* = {ε, b, bb, bbb, ...}.
The union of L1 and L2, (L1 ∪ L2), is the set {a, b}.
Now, let's analyze (L1 ∪ L2)∗, which represents the Kleene star operation applied to (L1 ∪ L2). (L1 ∪ L2)∗ consists of all possible combinations of 'a's and 'b's, including the empty string: (L1 ∪ L2)∗ = {ε, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, ...}.
Comparing (L1∗ ∪ L2∗) and (L1 ∪ L2)∗, we observe that they represent the same set of strings. Both sets include all possible combinations of 'a's and 'b's, including the empty string. Hence, we have (L1∗ ∪ L2∗) = (L1 ∪ L2)∗.
This example demonstrates that there exist languages L1 and L2 for which neither is a subset of the other, but their respective Kleene star unions result in the same set of strings.
To learn more about Kleene star operation: -brainly.com/question/12976788
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