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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.
1. Does above all''s properties can be used to prove a language regular? 2..which of the properties can be used to prove a language regular and which of these not? 3..Identify one
We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also
Construct a PDA that accepts { x#y | x, y in {a, b}* such that x ? y and xi = yi for some i, 1 = i = min(|x|, |y|) }. For your PDA to work correctly it will need to be non-determin
let G=(V,T,S,P) where V={a,b,A,B,S}, T={a,b},S the start symbol and P={S->Aba, A->BB, B->ab,AB->b} 1.show the derivation sentence for the string ababba 2. find a sentential form
automata of atm machine
write short notes on decidable and solvable problem
DEGENERATE OF THE INITIAL SOLUTION
Intuitively, closure of SL 2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a
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
When an FSA is deterministic the set of triples encoding its edges represents a relation that is functional in its ?rst and third components: for every q and σ there is exactly one
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