Lecture Notes in Computer Science 3472

s1
s3
a/0
b/0
a/1
s2
s4
3 State Verification 77
b/1
Fig. 3.5. The inference graph of the machine in Figure 3.3.
Proposition 3.6 (Inference [Nai97]). Let G be the transition graph of a Mealy
machine M. If state si is reachable from state sj in the inference graph GI ,and
si has a UIO sequence, then sj has a UIO sequence.
Proof. Let x be a UIO sequence for state si, andx ′ the sequence of inputs
along a path from sj to si in GI . The claim is that x ′ · x is a UIO sequence
for sj . 4 Suppose this is not the case, i.e., λ(sk , x ′ · x )=λ(sj , x ′ · x )forsome
state sk �= sj . We first show that the unique path with label x ′ from sk must
end in si. Ifitendedinsomeotherstatesl �= si, thenλ(sl, x )=λ(si, x ), since
λ(sk , x ′ · x )=λ(sj , x ′ · x ), but this is impossible, since x is a UIO sequence for
si.
Next, we show that in fact there can be no state sk �= sj such that λ(sk , x ′ ·
x )=λ(sj , x ′ · x ). Again assume that there is. Let sj = sj1, sj2,...,sjm = si and
sk = sk1, sk2,...,skm = si be the two paths with input label x ′ from sj and sk ,
respectively. Consider the smallest i such that sji and ski are identical. The edges
from sji−1 to sji and from ski−1 to sji in G must have the same label and end up
in the same state. Therefore, they are converging, and none of them belongs in
the inference graph GI . This means that si cannot be reachable from sj along a
path with input label x ′ and output label λ(sj , x ′ )inGI . This is a contradiction,
and we conclude that λ(sj , x ′ · x ) �= λ(sk , x ′ · x ) for all sk �= sj .Thusx ′ · x is a
UIO sequence for sj .
As an example, once we know that b is a UIO sequence for s3 in the machine
from Figure 3.3, we can use the inference graph in Figure 3.5 to conclude that
ab must be a UIO sequence for state s1.
The inference graph can also be used to obtain negative answers to the question
whether a state has a UIO sequence.
4 The dot operator for sequences (X · Y ) denotes concatenation.