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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.
This was one of the ?rst substantial theorems of Formal Language Theory. It's maybe not too surprising to us, as we have already seen a similar equivalence between LTO and SF. But
To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the
De?nition (Instantaneous Description) (for both DFAs and NFAs) An instantaneous description of A = (Q,Σ, δ, q 0 , F) , either a DFA or an NFA, is a pair h q ,w i ∈ Q×Σ*, where
. On July 1, 2010, Harris Co. issued 6,000 bonds at $1,000 each. The bonds paid interest semiannually at 5%. The bonds had a term of 20 years. At the time of issuance, the market r
Exercise: Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.
This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL 2 to discover properties of the recognizable languages.
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
mmmm
constract context free g ={ a^n b^m : m,n >=0 and n
The key thing about the Suffx Substitution Closure property is that it does not make any explicit reference to the automaton that recognizes the language. While the argument tha
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