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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. Nevertheless, this does not contradict the fact that every ?nite language is SLk, for some k. All it says is that the counterexamples that establish non-closure of SL under union and concatenation must involve non-?nite languages. (You should verify the fact that they do.)
a finite automata accepting strings over {a,b} ending in abbbba
The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular
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.
Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
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
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 .
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
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
turing machine
Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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