Here is the transition function of a simple, deterministic automaton with start state A and accepting state B:

	0	1
A	A	B
B	B	A

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.

Here, is the extended transition function of the automaton; that is, (A,w) is the state that the automaton is in after processing input string w. The proof of the statement 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.

B)

Reasoning about properties of deterministic finite automata, e.g., that if string s = yz, then (q,s) = ((q,y),z).

C)

Reasoning about properties of binary strings (strings of 0's and 1's), e.g., that every string is longer than any of its proper substrings.

Basis (|w| = 0):

(1)

w = because ________

(2)

(A,) = A because ________

(3)

has an even number of 1's because ________

Induction (|w| = n > 0)

(4)

There are two cases: (a) when w = x1 and (b) when w = x0 because ________

Case (a):

(5)

In case (a), w has an odd number of 1's if and only if x has an even number of 1's because ________

(6)

In case (a), (A,x) = A if and only if w has an odd number of 1's because ________

(7)

In case (a), (A,w) = B if and only if w has an odd number of 1's because ________

Case (b):

(8)

In case (b), w has an odd number of 1's if and only if x has an odd number of 1's because ________

(9)

In case (b), (A,x) = B if and only if w has an odd number of 1's because ________

(10)

In case (b), (A,w) = B if and only if w has an odd number of 1's because ________

		a)	(1) for reason A.
		b)	(6) for reason C.
		c)	(10) for reason C.
		d)	(9) for reason A.