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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.
Design a turing machine to compute x + y (x,y > 0) with x an y in unary, seperated by a # (descrition and genereal idea is needed ... no need for all TM moves)
write short notes on decidable and solvable problem
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
We will assume that the string has been augmented by marking the beginning and the end with the symbols ‘?' and ‘?' respectively and that these symbols do not occur in the input al
What is the arbwnememmsmdbdbfbfjmfksmjejfnfnfnnrndmnfjfjfnrnkrkfjfnfmkrjrbfbbfjfnfjruhrvrjkgktithhrbenfkiffnbr ki rnrjjdjrnrk bd n FBC..jcb?????????????????????????????????????????
The fact that regular languages are closed under Boolean operations simpli?es the process of establishing regularity of languages; in essence we can augment the regular operations
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Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
write grammer to produce all mathematical expressions in c.
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