We want to show that this automaton accepts exactly those strings with an odd number of 1's, or more formally: (A,w) = B if and only if w has an odd number of 1's.

5. Here is the transition function of a simple, deterministic automaton with start state A and accepting state B 0 1 AAB We want to show that this automaton accepts exactly those strings with an odd number of l's, or more formally: 6(A, w) B if d only if w has an odd number of l's w Here, S is the extended transition function of the automato that is, o(A, is the state that the automaton is in after processing input s w. The proof of the stalemenu above is an induction on the length of w, Below, we give the proof with reasons missing. You must give a reason for each step, and then demonstrate your understanding of the proof by classifying your reasons into the following three categories: A) Use of the inductive hypothesis. Reasoning about properties of deterministic finite automata, e.g., that if string Reasoning about proporties of binary strings (strings of 0's and l's e.g., that every string is longer than any of its proper substrings. Basis w 0) w because (ASE) A because E has an even number of 1's because duction (w n 0) There are two cases (a) when w X1 and (b) when w x00 because Case (a In case (a), w has an odd number of l's if and only ifx has an even number of l's because In case (a), OCA,x) A if and only if w has an odd number of 1's because In case (a), 0(A,w) Bif and only if w has an odd number of l's because Case (b In case (b), w has an odd number of 1's if and only if x as an odd number of l's because In case (b)T 50A,x) Bif and only if w has an odd number of 1's because (10) In case (b), DIA, Bif and only if w has an odd number of l's because a) (7) for reason C b) (1) for reason A c) (8) for reason A d) (8) for reason C