sufficient to demonstrate that mA is accessible in some finite number of transitions from
every state n e S' SA'.
Let iA E S be the bit-string represented in mA, let n e S' SA' and let i, e S be
selected such that n(i,) > 0 and H(i1, IA) H(i, iA) for all i represented in n. Then, two
cases must be examined. In case (1), ii = iA while i t iA for case (2).
If ii = iA, it follows from Eq. 7.7 and the construction of Q' that
Q'(mA n)= lim Pi(mA I n) = lim P2(mA I n)
a 0+ ta --0+
[lim P2iA In)]M
= P(iA I n)M = P(i, I n)M > 0
and consequently mA is accessible from n in 1 transition. Otherwise
3i2e S(i,) 3 H(i2, A) = H(i,,A) 1
and further if n, e SA' is the one-operator absorbing state defined by the condition
n,(ii) = M while n,2 e S(n,)' is the adjacent nonabsorbing state defined by
nl2(i) = M 1, n2(i2) = 1, then from Eq. 7.7 and the construction of Q'
Q'(nl2 1 n) = lim P2(n2 I n)+ lim P2(, n)
0W L, 0+
a -0 Ia -o
1
= P,(n2 |n)+-P1(n, In)
L
= P,(n, n)
= [P,(i I n)]M > 0.
Thus, nl2 is accessible from n in one transition. If i2 = iA, then by the case (1) argument mA
is accessible in one additional transition. Otherwise, the case (2) argument is repeated for
some
i3 E S(i2) 3 H(i3, iA)= H(i2 ) 1 = H(i,, iA) 2.