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The fact that SL2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem
L1 ∩ L2 =
We know that the intersection of SL2 languages is also SL2. If the complement of SL2 languages was also necessarily SL2, then would be SL2 contradicting the fact that there are SL2 languages whose union are not SL2.
Lemma The class of strictly 2-local languages is not closed under complement .
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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Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r
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One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
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