Myhill-Nerode theorem
Let be a language on the finite alphabet
and let be its Nerode equivalence.
The following are equivalent
![]()
.
-
1.
is recognized by a deterministic finite automaton.
-
2.
is finite.
Moreover, the number of classes of is the smallest number of states of a DFA recognizing .
Proof.
First, suppose
where is the empty word on .
Construct a DFA
(called the Nerode automaton for )
with defined by
| (1) |
and
| (2) |
Then is well defined because implies . It is also straightforward that recognizes .
On the other hand, let be a DFA that recognizes . Extend to by putting and for every , , . Define as
| (3) |
Then is well defined.
In fact, suppose :
then either ,
or there are such that .
But in the latter case,
for any ,
hence since recognizes .
Finally, for any we have
so every class of has a preimage![]()
according to ;
consequently, .
| Title | Myhill-Nerode theorem |
|---|---|
| Canonical name | MyhillNerodeTheorem |
| Date of creation | 2013-03-22 18:52:13 |
| Last modified on | 2013-03-22 18:52:13 |
| Owner | Ziosilvio (18733) |
| Last modified by | Ziosilvio (18733) |
| Numerical id | 14 |
| Author | Ziosilvio (18733) |
| Entry type | Theorem |
| Classification | msc 68Q70 |
| Classification | msc 20M35 |
| Synonym | Myhill-Nerode theorem for languages |