
P1: IML
MOBK015-05 MOBK015-Lynch.cls April 1, 2006 17:2
68 THEORY OF TIMED I/O AUTOMATA
In the rest of this section, we sometimes speak of variables, states, and traces of a TA
with bounds. If A = (B, C, l, u) is a TA with bounds, variables, states, and traces of A refer to,
respectively, the states and the traces of the underlying automaton B.
Theorem 5.18 Suppose A is a TA with bounds. Then
traces
Extend(A)
⊆ traces
A
.
Proof: Let F : Q
→ Q be defined as follows: F(x) = x X, where X is the set of internal
variables of A. It is easy to check that F is a refinement from Extend(A)toA. By Theorem 4.27
and Corollary 4.23, we conclude that
traces
Extend(A)
⊆ traces
A
.
Lemma 5.19 Suppose that A = (B, C, l, u) is a TA with bounds. For any reachable state x of
Extend(A), if C is enabled in x XinA, then x(
last) ≤ x(now ) + u.
Proof: Consider a closed execution α of Extend(A). Using Axioms T1 and T2 for trajectories,
we can write α as a concatenation of closed execution fragments α
0
α
1
···
α
k
, where α
0
is a point trajectory and each α
i
for i ≥ 1 is either a trajectory or a discrete action surrounded
by two point trajectories such that for all 0 ≤ i ≤ k − 1, α
i
.lstate = α
i+1
.fstate. We prove the
invariant by induction on the length k of the sequence of execution fragments.
For the base case, suppose that C is enabled in α
0
.fstate X. Since α is an execution, we
know that α
0
.fstate is a start state of Extend(A). By definition of Extend(A), α
0
.fstate(last) =
u. Since α
0
.fstate(now) = 0, α
0
.fstate(last) ≤ α
0
.fstate(now) + u, as required.
For the inductive step, we assume that the property is true for the sequence α
0
α
1
···
α
k
, and show that it is true in the sequence α
k+1
in α
0
α
1
···
α
k
α
k+1
.
There are two cases to consider depending on whether α
k+1
is a discrete action surrounded by
two point trajectories or a trajectory.
1. α
k+1
is an action a surrounded by two point trajectories ℘(y) and ℘(y
). Suppose that
C is enabled in y
X in A. There are two subcases to consider:
a) C is enabled in y X and a /∈ C. Then, y
(last) = y(last) and y
(now) = y(now).
By inductive hypothesis, y(
last) ≤ y(now ) + u. Therefore, y
(last) ≤ y
(now) + u,
as needed.
b) C is disabled in y X or a ∈ C. Then, by definition of Extend(A), y
(last) =
y
(now) + u, which suffices.
2. α
k+1
is a trajectory. Suppose that C is enabled in α
k+1
.lstate X in A. There are two
subcases to consider:
a) C is enabled in α
k+1
.fstate X in A. By inductive hypothesis α
k+1
.fstate(last) ≤
α
k+1
.fstate(now) + u. By the well-formedness assumption, we know that C must be