Final answer:
The language L={w∈{0,1}∗:w=0m,m≥0} is a regular language, while the language L={w∈{0,1}∗:w=0m1m,m≥0} is not. The input 0111 is accepted by the DFA in question 5, and a decomposition that satisfies the conditions of the Pumping Lemma is 0111=ryz where r=0, y=1, and z=11.
Step-by-step explanation:
(a) Let's examine the language L={w∈{0,1}∗:w=0m,m≥0}. This language consists of all strings of the form 0 followed by zero or more 0s. Since this language can be represented with a simple regular expression (0*), it is a regular language. Therefore, the statement (a) is TRUE.
(b) Now let's consider the language L={w∈{0,1}∗:w=0m1m,m≥0}. This language consists of all strings of the form 0 followed by some number of 1s, where the number of 0s and 1s are equal. This language cannot be represented by a regular expression since there is no way to keep track of the count of 0s and 1s. Therefore, the statement (b) is FALSE.
(a) To show that w=0111∈L(A), we need to trace the path of the DFA for the input 0111. Starting from the initial state, we transition to the state labeled 0 on the first input 0. Then, we transition to the state labeled 1 on the second input 1. Finally, we transition to the state labeled 2 on the last two inputs 1. Since the DFA ends in an accepting state, the input 0111 is accepted and therefore w=0111∈L(A).
(b) To find a decomposition 0111=ryz that satisfies the conditions of the Pumping Lemma, we can choose r=0, y=1, and z=11. Therefore, we have xy^2z=0111 which is in L(A) for all k≥0. This decomposition satisfies the conditions of the Pumping Lemma.