Introduction to Computational Linguistics

12. Finite State Automata 41
The proof is by induction on the length of ⃗u. It is now easy to see that L(A × B) =
L(A)∩L(B). For ⃗u ∈ L(A×B) if and only if 〈i A 0 , iB 0 〉 ⃗u → 〈x 0 , x 1 〉, which is equivalent
to i A 0
⃗u
→ x 0 and i B 0
⃗u
→ x 1 . The latter is nothing but ⃗u ∈ L(A) and ⃗u ∈ L(B).
Now, assume that A and B are total. Define the set G := F A × Q B ∪ Q A ×
F B . Make G the accepting set. Then the language accepted by this automaton is
L(A) ∪ L(B) = s ∪ t. For suppose u is such that 〈i A 0 , iB 0 〉 ⃗u → q ∈ G. Then either
q = 〈q 0 , y〉 with q 0 ∈ F A and then ⃗u ∈ L(A), or q = 〈x, q 1 〉 with q 1 ∈ F B . Then
⃗u ∈ L(B). The converse is also easy.
Here is another method.
Definition 7 Let L ⊆ A ∗ be a language. Then put
(119) L p = {⃗x : there is ⃗y such that ⃗x ⌢ ⃗y ∈ L}
Let s be a regular term. We define the prefix closure as follows.
(120a)
(120b)
(120c)
(120d)
(120e)
(120f)
0 † := ∅
a † := {ε, a}
ε † := {ε}
(s ∪ t) † := s † ∪ t †
(st) † := {su : u ∈ t † } ∪ s †
(s ∗ ) † := s † ∪ {s ∗ u : u ∈ s † }
Notice the following.
Lemma 8 ⃗x is a prefix of some string from s iff ⃗x ∈ L(t) for some t ∈ s † . Hence,
⋃
(121) L(s) p = L(t)
t∈s †
Proof. Suppose that t ∈ s † . We show that L(t) is the union of all L(u) where
u ∈ s † . The proof is by induction on s. The case s = ∅ and s = ε are actually easy.
Next, let s = a where a ∈ A. Then t = a or t = ε and the claim is verified directly.