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Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff ≡L partitions Σ* into ?nitely many Nerode equivalence classes.
Proof: For the "only if" direction (that every recognizable language has ?nitely many Nerode equivalence classes) observe that L ∈ Recog iff L = L(A) for some DFA A and that if δ(q0,w) = δ(q0, u) (i.e., if the path from the start state labeled w and that labeled u end up at the same state) then w ≡L u. This is a consequence of the fact that the state ˆ δ(q0,w) encodes all the information the automaton remembers about the string w. If v extends w to wv ∈ L(A) then v is the label of a path to an accepting state from δ(q0,w). Since this is the same state as δ(q0, u) the same path witnesses that uv ∈ L. Similarly, if the path leads one to a non-accepting state then it must necessarily lead the other to the same state. The automaton has no way of distinguishing two strings that lead to the same state and, consequently, the language it recognizes cannot distinguish them. Since A is deterministic, every string in Σ* labels a path leading to some state, hence the equivalence classes corresponding to the states partition Σ*. Since the automaton has ?nitely many states, it distinguishes ?nitely many equivalence classes.
Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
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The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that
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