prove that (A ∪ B)* = A*(BA*)*
To prove that (A ∪ B)* = A*(BA*)*, we need to show that both sides of the equation contain the same elements.
To start, let's first understand the notation used:
- A ∪ B represents the union of sets A and B, which contains all the elements that are in set A or set B (or both).
- * denotes the Kleene star operator, which represents the concatenation of zero or more occurrences of the given set or language.
- A* represents any number of concatenations (including zero occurrences) of elements from set A.
Now, let's analyze each side of the equation separately:
Left-hand side (LHS): (A ∪ B)*
The Kleene star of the union of sets A and B represents any number of concatenations (including zero occurrences) of elements from either A or B. This means that the LHS can contain elements from A, elements from B, or elements that are common to both sets.
Right-hand side (RHS): A*(BA*)*
Let's break down the RHS step by step:
- A* represents any number of concatenations (including zero occurrences) of elements from set A.
- BA* represents the concatenation of an element from B followed by any number of occurrences of elements from A.
- Finally, (BA*)* represents any number of concatenations (including zero occurrences) of elements that start with an element from B followed by any number of occurrences of elements from A.
Now, let's compare the two sides:
For an element to exist on the LHS, it has to be a concatenation of zero or more elements from either A or B. If it contains only elements from A, we can match it on the RHS by considering A* since A* allows any number of concatenations of elements from A (including zero occurrences). Similarly, if it contains elements from B, we can match it on the RHS by considering the part BA* because BA* allows any number of occurrences of elements from A, preceded by one element from B.
Therefore, we can conclude that any element on the LHS is also contained on the RHS, and vice versa. Hence, (A ∪ B)* = A*(BA*)*.