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Theorem The class of recognizable languages is closed under Boolean operations.
The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a given string is in either or both of any pair of recognizable languages. We can modify the construction for other Boolean operations simply by selecting the appropriate set of accepting states:
• Union: Let F′
= {(q, p) | q ∈ F1 or p ∈ F2}. Then L(A′ ) = L1 ∪ L2.
• Relative complement: Let F′ = F1 × (Q2 - F2). Then L(A′ ) = L1 -L2.
• Complement: Let L1 = Σ* and use the construction for relative complement.
Theorem The class of ?nite languages is a proper subclass of SL. Note that the class of ?nite languages is closed under union and concatenation but SL is not closed under either. N
The class of Strictly Local Languages (in general) is closed under • intersection but is not closed under • union • complement • concatenation • Kleene- and positive
Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complica
De?nition Instantaneous Description of an FSA: An instantaneous description (ID) of a FSA A = (Q,Σ, T, q 0 , F) is a pair (q,w) ∈ Q×Σ* , where q the current state and w is the p
Application of the general suffix substitution closure theorem is slightly more complicated than application of the specific k-local versions. In the specific versions, all we had
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
Different types of applications and numerous programming languages have been developed to make easy the task of writing programs. The assortment of programming languages shows, dif
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
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